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PROFESSOR: Today,
we continue-- we're

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going to do two things today.

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One is protein folding and the
other is interlocked chains.

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So protein folding,
we're continuing

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on where we left off last
time, which was essentially

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looking at the mechanics, the
mechanical models of protein

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folding, which was
fixed angle chains.

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But basically ignoring
all the forces that

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go into actual protein folding.

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Today, we're going
to look at some

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of the very theoretical
work on what

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forces cause proteins
to fold and what

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you can say about algorithms and
complexities of those problems.

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So obviously, a lot of people
work on protein folding.

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On the practical side,
people look a lot

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at defining energy functions
that you minimize in order

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to find what we believe would
be the right protein folding.

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There's a lot of energy
functions around,

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and the key energies that--
forces people look at

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are torsion angles potentials,
van der Waal interactions,

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hydrogen bonds, hydrophobicity,
secondary structure propensity,

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and paralyze specific
interactions.

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And then they take some weighted
combination of all those things

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and try to find
one that matches up

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with what we see in real life.

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The things we see
in real life, we

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get by crystallizing
proteins, which

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can take a really long time,
like years of research.

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Then you get one 3D image.

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And that's what protein data
bank collects is all those 3D

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images.

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People try to train against
that data, but I say,

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we really don't know
what the right objective

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is for real protein folding.

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Given that, because
I'm a theoretician,

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we're going to look at
a super simple model

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that just models one of those
features that I mentioned,

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which is hydrophobicity.

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Hydrophobia is fear of water.

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And remember we have
this back on the protein,

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the various amino acids hanging
off, some of those amino acids

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are hydrophobic, meaning
they like to hide from water.

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Now usually, your proteins live
in a big bath of fluid, mostly

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water, and so they're trying
to hide from the surroundings,

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try to cluster on the
inside of the folded shape,

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whereas other hydrophilic amino
acids that like water, they

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tend to go around the outside.

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You can't always
do this, but you

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try to minimize the amount to
which the hydrophobic guys are

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on the outside and maximize
the hydrophilic guys are

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on the outside.

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In the HP model, there are
hydrophobic and hydrophilic

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guys.

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And actually, we'll
call h the hydrophobic

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and P the hydrophilic.

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It's very confusing
because both of them

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have h and p in the acronym,
but p also apparently stands

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for polar, polar opposite
of the hydrophobic guys.

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Anyway, we just have
these two kinds of nodes,

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and we just try
to model that one

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feature, plus the bonds
that link together,

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the backbone of the chain.

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And so we're going
to model a chain

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as just a sequence of nodes.

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We're going to fold
that chain on a lattice.

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Something like this.

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Is it five?

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Yeah.

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And different of these nodes
could be marked h or p.

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And we usually could use
two colors to denote that.

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In this case, I made this guy
h, this guy h, this guy h.

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The score of a fold--
so this is a folding.

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We're always going to
fold on the square lattice

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today although people have
looked at other lattices,

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in particular 3D cubic
lattice and a little bit

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on the triangular lattice.

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I'll mention some reasons
why the square lattice is

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a little bit artificial,
but in general, conjectures,

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it doesn't really matter
which lattice you fold on.

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But you really want
to follow lattices

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because we're going to
define the score in terms

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of how many adjacent H nodes
there are on the lattice.

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Score is the number of
call them H-H bonds.

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The intuition is if you have
two hydrophobic nodes that

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are next to each other
that means they're not

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next to the boundary,
and so they're happier.

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There's a more formal
argument of that,

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but this is an easy
definition of score.

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You want to maximize
the number of H-H nodes

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that are adjacent
on the lattice.

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Usually, there are some that
are adjacent along the chain.

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Those we don't care about
because they will always

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be adjacent.

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You have to fold this thing
so that consecutive nodes

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in the chain appear
adjacent in the lattice.

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So we usually ignore what I
would call H-H edges which

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are four, so you can
count them or not.

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It doesn't really matter.

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We want to maximize the
number of these guys.

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For this chain, that's probably
the maximum you can get.

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In general, a typical node
has two edge neighbors,

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and it has two other neighbors.

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And so at most, every
guy has-- every H node

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has two bonds next
to it, which means

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on average at most
one bond per H node

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because we're double counting.

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So that's the model.

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The optimal folding
has the maximum score.

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And I'm going to
talk about the theory

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of this model in a moment.

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Some practice you might
say is this picture.

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This is heuristic or local
optimization of maximizing.

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Here, red nodes are H nodes that
will be throughout the slides.

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Red nodes are H nodes.

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Blue nodes are P nodes.

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That's a universal standard.

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You can see, in this
case conveniently--

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it may look like some random
pattern of reds and blues,

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but you end up with all the
blues pretty much covering

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the outside, and so the
reds cluster together,

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and there's lots of H-H bonds.

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They are not drawn here,
but-- So that's nice.

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And this looks like
a typical protein

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in that it globs together.

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But of course, it's not
capturing every feature

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of protein folding, just
trying to capture one of them.

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But pretending this
is the whole story,

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what can you say about optimal
folding in the HP model?

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Well sadly, it's NP-hard.

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So to find the actual
optimal folding of a given

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string of red and
blue nodes is NP-hard.

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It's NP-hard in the
3D cubic lattice.

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That was by two MIT professors,
Bonnie Berger and Tom Leighton

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And then it was proved
NP-hard hard in two dimensions

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by whole bunch of people,
I guess, from Berkeley.

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So that's bad news.

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And to me, that's a sign
that nature probably

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does not find the
optimal folding.

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There are a couple
possibilities here.

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Maybe they're extra forces
beyond hydrophobicity

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that make it easier to
find the optimal solution

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or maybe nature's just
finding a local optimum, not

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a global optimum.

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Or maybe the proteins that
exist in real life that

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have been evolved are designed
so that local optima equal

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global optima.

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That's somewhat related.

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Somehow nature gets around this
NP-hardness, I would think,

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because I believe
nature's a computer.

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Maybe not everyone does.

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But it should be bounded by
polynomial time computation.

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So unless P equals NP-- that's
another possibility I guess.

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It's unlikely that, to
me, nature is doing this.

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But maybe it's
finding local optimum.

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That seems to do pretty well
in this particular example.

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AUDIENCE: Hey Erik?

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PROFESSOR: Yeah.?

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AUDIENCE: Are there
approximation algorithms?

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PROFESSOR: Good question.

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Are there approximation
algorithms,

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that's exactly where
I wanted to go.

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There are lots of constant
factor approximations,

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and I will show you one of them.

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So I'm not going to show
you these hardness proofs

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because they're a
bit complicated.

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I don't have free
diagrams of them.

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But the constant factor
approximations, I do.

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Where are we going?

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Here.

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And so, in particular, the
best approximation that's known

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is a 1/3 approximation.

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This is by MIT Ph.D.
student [INAUDIBLE],

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and from several years
ago-- eight years ago.

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I believe it's still the best.

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It's pretty simple,
but quite elegant,

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looks like protein folding.

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So this is going to be
within a factor of three

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of the best you could hope for.

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We write 1/3 because we're
always getting fewer bonds

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than you might want in
a maximization problem.

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Some people call this
three approximation.

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I call it one of the other
depending on the day the week.

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So how could we achieve
1/3 approximation?

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Well, before I get to this
image, let me tell you.

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In general, when-- we
haven't talked too much

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about approximation
algorithms in this class,

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but a typical technique in
approximation algorithms is you

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get some bound on what
the optimal might be.

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I already told you there's at
most one bond for each H node.

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And that is indeed-- that's
roughly what we'll say.

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Just because of the
degree argument,

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every guy has bond
degree at most two.

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So that means, at most,
one edge per H node.

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But we can be a little more
precise on the square grid

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because there are
two types of nodes.

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If you checkerboard
color the square grid,

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there's even nodes
and odd nodes.

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So in fact, we claim that
the optimal solution,

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which we call OPT, is
always at most twice the min

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of the number of even H's
and the number of odd H's.

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This is slightly smaller
than the number of H's.

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If a number of H's
and not H's are equal,

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then this will just
be the number of H's.

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But if one is smaller
than the other,

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this will be a
little bit smaller

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or it could be a lot smaller.

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They're saying, no even H's.

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The point is if you
have an even parity H,

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and it somehow has a
bond with some other H,

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well obviously, that h has
different parity from this one.

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So only even and odd H's
could possibly bond together.

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And then this is pretty obvious.

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Let's see if I can formalize it.

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So you have every bond
defines and even and an odd.

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Every even and odd can
only get hit up to twice.

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If there's more evens and odds,
it's not going to help you.

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In the best case, you pair
them up or alternate even odd,

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even odd, even odd.

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So you'll never be able to
use more evens than odds,

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maybe one more, not really.

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And I think the best case is
you do a cycle of even odd,

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even odd, even odd.

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Then, there's an even
number of them-- each,

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an equal number of each.

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And so you throw
away the excess.

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You end up with
a min of the two.

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But you get a factor of two
because every even guy has

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two incident bonds
in the best case,

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and then you're
not double counting

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because you count this even guy,
maybe you count this even guy,

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and they have disjoint bonds.

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So it's twice the number
of evens or number

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of odds, whichever is smaller.

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So this is useful.

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It's just what is the best
case you could hope for OPT?

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Now, what this 1/3
approximation will get

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is at least 1/3 this amount.

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If you know you get at
least 1/3 this amount,

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you know you get at
least 1/3 the optimal

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because this is an upper
bound on the optimal.

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So that's with the
algorithm will do.

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And so the general
idea of the algorithm

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is to proceed in stages.

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At a typical stage--
So you pick some edge.

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I don't think it'll
matter too much.

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Pick some edge of the
chain, and then you

247
00:12:47,830 --> 00:12:50,220
start working your way down.

248
00:12:50,220 --> 00:12:54,920
And general idea is you can
make big loops off to the sides

249
00:12:54,920 --> 00:13:00,780
so that when you get back to
the center, these are H nodes.

250
00:13:00,780 --> 00:13:02,030
So you get a bond there.

251
00:13:02,030 --> 00:13:03,380
That's the general idea.

252
00:13:03,380 --> 00:13:07,320
With that basic idea, you
probably get about 1/4

253
00:13:07,320 --> 00:13:08,330
of this amount.

254
00:13:08,330 --> 00:13:12,310
But we want to get 1/3,
which is a little better.

255
00:13:12,310 --> 00:13:15,250
I won't try to
analyze that formally.

256
00:13:15,250 --> 00:13:17,000
But there's a lot
of flexibility here

257
00:13:17,000 --> 00:13:19,640
when we-- how long
we make this branch.

258
00:13:19,640 --> 00:13:21,450
In order to make
this work out, what

259
00:13:21,450 --> 00:13:29,110
I like is that the left branch
has a lot of-- pick one-- a lot

260
00:13:29,110 --> 00:13:33,900
of even H's, and the right
branch has a lot of odd H's,

261
00:13:33,900 --> 00:13:36,240
or vice versa
because then we'll be

262
00:13:36,240 --> 00:13:37,790
able to make lots
of the even guys

263
00:13:37,790 --> 00:13:39,270
match up with the odd guys.

264
00:13:39,270 --> 00:13:41,060
Here, we're fighting the parity.

265
00:13:41,060 --> 00:13:42,700
And parity is one
of the reasons why

266
00:13:42,700 --> 00:13:44,992
I think the square grid is
a little bit artificial.

267
00:13:44,992 --> 00:13:47,450
Let's say with the triangular
grid, you wouldn't have that.

268
00:13:47,450 --> 00:13:50,490
There aren't just two classes.

269
00:13:50,490 --> 00:13:51,970
But here, we can
get around parity

270
00:13:51,970 --> 00:13:55,310
by finding a break
point, this edge,

271
00:13:55,310 --> 00:13:58,500
so that the number of
even guys in the left

272
00:13:58,500 --> 00:14:01,350
is about the number of
odd guys on the right.

273
00:14:01,350 --> 00:14:03,990
I won't prove that, but it's
just an intermediate value

274
00:14:03,990 --> 00:14:05,230
argument.

275
00:14:05,230 --> 00:14:08,020
See, you get that as you
walk counterclockwise

276
00:14:08,020 --> 00:14:11,040
along the chain, you
have more even guys,

277
00:14:11,040 --> 00:14:15,360
and walk clockwise around the
chain, you get more odd guys.

278
00:14:15,360 --> 00:14:18,080
Once you have that set up,
you fall into four cases.

279
00:14:22,450 --> 00:14:25,802
So we're supposing at the
top-- let me get to a case,

280
00:14:25,802 --> 00:14:26,510
that I can reach.

281
00:14:26,510 --> 00:14:29,170
Here's a typical case.

282
00:14:29,170 --> 00:14:31,630
Suppose you just
succeeded in making

283
00:14:31,630 --> 00:14:32,860
two H nodes join together.

284
00:14:32,860 --> 00:14:34,850
Maybe that was your
first edge, whatever.

285
00:14:34,850 --> 00:14:38,030
So it might not be H-H. But
now, we're proceeding down,

286
00:14:38,030 --> 00:14:40,795
and we want to align
a bunch of H nodes.

287
00:14:40,795 --> 00:14:43,170
And we're going to do better
than the picture I drew over

288
00:14:43,170 --> 00:14:45,590
here, which was just getting
two H nodes to come together.

289
00:14:45,590 --> 00:14:48,020
I actually want to
get four H nodes

290
00:14:48,020 --> 00:14:49,850
to come together
with three bonds.

291
00:14:49,850 --> 00:14:52,900
That's what I will always
achieve, at least that.

292
00:14:52,900 --> 00:14:53,590
Is that right?

293
00:14:53,590 --> 00:14:55,090
Actually, no, this
is the good case.

294
00:14:55,090 --> 00:14:57,970
I won't always achieve quite
that well, quite that good.

295
00:14:57,970 --> 00:14:59,060
But here's the idea.

296
00:14:59,060 --> 00:15:04,720
Suppose there's some reasonably
large distance between this H

297
00:15:04,720 --> 00:15:06,560
node and the next one.

298
00:15:06,560 --> 00:15:10,960
I'm going to go down two steps
and then go around however

299
00:15:10,960 --> 00:15:15,120
long it takes so that when I
come back, I get to an H node,

300
00:15:15,120 --> 00:15:16,220
and same on the left side.

301
00:15:16,220 --> 00:15:18,136
So these two chains may
have different lengths

302
00:15:18,136 --> 00:15:20,109
depending on when
the next H node is.

303
00:15:20,109 --> 00:15:21,900
Now, when I say the
next H node, I actually

304
00:15:21,900 --> 00:15:23,610
mean the next odd
H node on this side

305
00:15:23,610 --> 00:15:27,210
and the next even H node on this
side so that they will line up.

306
00:15:27,210 --> 00:15:29,870
This is assuming that
the distance from this H

307
00:15:29,870 --> 00:15:32,190
node to the next one
is greater than 1,

308
00:15:32,190 --> 00:15:34,650
by parity that means
it's at least three.

309
00:15:34,650 --> 00:15:37,220
So if it's three, it's
going to be one, two, three.

310
00:15:37,220 --> 00:15:40,000
And that's it.

311
00:15:40,000 --> 00:15:41,276
And so go straight down.

312
00:15:41,276 --> 00:15:42,650
If it's more than
three, you just

313
00:15:42,650 --> 00:15:44,983
divide that distance in half,
and you go back and forth.

314
00:15:44,983 --> 00:15:47,190
And this sets the parity right.

315
00:15:47,190 --> 00:15:49,270
So that's the set up.

316
00:15:49,270 --> 00:15:51,250
Now, we have these two H nodes.

317
00:15:51,250 --> 00:15:52,980
Now, when are the next H nodes?

318
00:15:52,980 --> 00:15:55,440
Well, there's at
least one blue node

319
00:15:55,440 --> 00:15:58,960
in between because of parity
because we're looking again

320
00:15:58,960 --> 00:16:00,360
at odd H nodes on
the right side,

321
00:16:00,360 --> 00:16:02,950
even H nodes on the left side.

322
00:16:02,950 --> 00:16:04,680
So maybe it's just
one blue node,

323
00:16:04,680 --> 00:16:08,092
and we would go like this, and
here we would go like that.

324
00:16:08,092 --> 00:16:09,550
Or if it's more
than one blue node,

325
00:16:09,550 --> 00:16:11,215
we just lay them out like this.

326
00:16:11,215 --> 00:16:12,590
Goes to the
appropriate distance,

327
00:16:12,590 --> 00:16:14,940
so that when we come
back, we get a red.

328
00:16:14,940 --> 00:16:16,670
So we have two red H nodes here.

329
00:16:16,670 --> 00:16:19,500
I want to make two red H
nodes here no matter what.

330
00:16:19,500 --> 00:16:21,940
If it's just gap of one,
we would go straight, make

331
00:16:21,940 --> 00:16:25,650
a corner, otherwise,
we do these big loops.

332
00:16:25,650 --> 00:16:27,570
And the result is I
get these four H nodes

333
00:16:27,570 --> 00:16:32,810
in this pattern, just three
bonds for four H nodes.

334
00:16:32,810 --> 00:16:35,320
And before we analyze
what exactly that means,

335
00:16:35,320 --> 00:16:37,610
let me just tell you
about the other cases.

336
00:16:37,610 --> 00:16:41,880
So this is the case when there
is more than one blue node

337
00:16:41,880 --> 00:16:44,770
in between two H nodes.

338
00:16:44,770 --> 00:16:47,040
The other rows and
columns are when

339
00:16:47,040 --> 00:16:48,600
the left side or
the right side has

340
00:16:48,600 --> 00:16:51,780
just one blue node
from that red guy.

341
00:16:51,780 --> 00:16:53,540
So let's go to this
one for example.

342
00:16:53,540 --> 00:16:56,570
Here the left side has only one
blue node to the next red guy.

343
00:16:56,570 --> 00:16:58,600
The right side has
at least three.

344
00:16:58,600 --> 00:17:02,460
So then we'd go one, two,
three, and we get there.

345
00:17:02,460 --> 00:17:04,196
If it's a more than
three, then we go

346
00:17:04,196 --> 00:17:05,846
and we make that big loop.

347
00:17:05,846 --> 00:17:07,220
Now we have these
two read nodes.

348
00:17:07,220 --> 00:17:09,069
My goal now is to
put a red node here.

349
00:17:09,069 --> 00:17:13,849
If this chain only has one blue
node, I would make a corner.

350
00:17:13,849 --> 00:17:19,260
Otherwise, I'd loop around, and
then I go onto the next step.

351
00:17:19,260 --> 00:17:21,890
Here, I only got two
bonds for three nodes,

352
00:17:21,890 --> 00:17:24,860
so it's actually a
little worse in ratio.

353
00:17:24,860 --> 00:17:26,609
In the case where
they're both length one,

354
00:17:26,609 --> 00:17:29,170
I just go straight down,
make those two H nodes,

355
00:17:29,170 --> 00:17:31,310
and then I use the
same trick as over here

356
00:17:31,310 --> 00:17:34,320
to get that nice
zig-zag pattern.

357
00:17:34,320 --> 00:17:37,620
And this case is symmetric
to this one, just flipped.

358
00:17:40,510 --> 00:17:50,460
So in all cases, I
didn't skip anything.

359
00:17:50,460 --> 00:17:53,750
I did all of the even H
nodes on the right side.

360
00:17:53,750 --> 00:17:56,080
I did all of the odd nodes,
H nodes, on the left side.

361
00:17:59,570 --> 00:18:00,574
So I did skip something.

362
00:18:00,574 --> 00:18:02,490
I skipped all of the odd
guys on the left side

363
00:18:02,490 --> 00:18:04,470
and all the even guys
on the right side.

364
00:18:04,470 --> 00:18:07,400
So I skipped half
of the H nodes.

365
00:18:07,400 --> 00:18:12,520
So I get at least two
bonds-- I'm calling them

366
00:18:12,520 --> 00:18:25,830
H-H bonds-- for every
three interesting H nodes.

367
00:18:28,600 --> 00:18:30,660
By interesting, I just
mean even on the left,

368
00:18:30,660 --> 00:18:36,500
odd on the right, which is
really for every six H nodes.

369
00:18:36,500 --> 00:18:41,440
And so we get a factor of
2/6, also known as 1/3.

370
00:18:46,200 --> 00:18:50,340
Really here I mean
6 even H nodes.

371
00:18:50,340 --> 00:18:51,610
Let's say.

372
00:18:51,610 --> 00:18:54,340
We used the same number
of even and odds.

373
00:18:54,340 --> 00:18:57,340
And we had-- Is that right?

374
00:18:57,340 --> 00:18:57,840
Sorry.

375
00:18:57,840 --> 00:19:00,655
No, I guess I actually
made H nodes here.

376
00:19:00,655 --> 00:19:02,780
I always get a little
confused here because there's

377
00:19:02,780 --> 00:19:04,363
a factor of two, a
factor of two here.

378
00:19:04,363 --> 00:19:07,600
But in the end we get 1/3 bound.

379
00:19:11,030 --> 00:19:13,539
So that's the
approximation algorithm,

380
00:19:13,539 --> 00:19:14,580
and that's the best node.

381
00:19:14,580 --> 00:19:15,455
It looks pretty nice.

382
00:19:15,455 --> 00:19:19,860
It's like beta sheets, if
you've seen proteins folding.

383
00:19:19,860 --> 00:19:22,434
it depends, of course, on
the spacing of the H nodes,

384
00:19:22,434 --> 00:19:24,600
but you get-- in particular,
you get all the H nodes

385
00:19:24,600 --> 00:19:26,090
on the inside pretty much.

386
00:19:26,090 --> 00:19:29,560
I guess here is one exposed end.

387
00:19:29,560 --> 00:19:31,450
And here there's
one exposed end.

388
00:19:31,450 --> 00:19:33,790
These two cases, we're
doing especially well.

389
00:19:33,790 --> 00:19:37,424
Maybe in real proteins, the
gaps are always more than one.

390
00:19:37,424 --> 00:19:38,340
I don't actually know.

391
00:19:40,659 --> 00:19:42,450
I think in reality
there's different levels

392
00:19:42,450 --> 00:19:46,080
of hydrophobicity, so that's
maybe not so well defined.

393
00:19:46,080 --> 00:19:48,154
But any questions?

394
00:19:48,154 --> 00:19:52,627
AUDIENCE: Is the exposed end
considered the [INAUDIBLE]

395
00:19:52,627 --> 00:19:54,194
above the dashed line or is it--

396
00:19:54,194 --> 00:19:54,860
PROFESSOR: Yeah.

397
00:19:54,860 --> 00:19:56,420
When I say exposed
end, I mean here.

398
00:19:56,420 --> 00:20:00,267
There's an adjacency
to the outside water.

399
00:20:00,267 --> 00:20:02,350
Now of course, there might
actually be nodes here.

400
00:20:02,350 --> 00:20:03,974
Then, that would not
be an exposed end.

401
00:20:03,974 --> 00:20:05,720
But if this just
went straight up,

402
00:20:05,720 --> 00:20:08,610
which it might in this
picture, then that

403
00:20:08,610 --> 00:20:09,660
would be an exposed end.

404
00:20:09,660 --> 00:20:11,970
I guess that might also be.

405
00:20:11,970 --> 00:20:13,300
No, probably not.

406
00:20:13,300 --> 00:20:15,300
Actually, interesting.

407
00:20:15,300 --> 00:20:16,731
If these two guys
are red, there's

408
00:20:16,731 --> 00:20:18,730
always a blue node to the
left and right of them

409
00:20:18,730 --> 00:20:20,710
in this construction,
so actually, I

410
00:20:20,710 --> 00:20:23,580
think that would not be exposed.

411
00:20:23,580 --> 00:20:25,520
That's cool.

412
00:20:25,520 --> 00:20:27,660
So actually, I think
there are no exposed

413
00:20:27,660 --> 00:20:29,760
ends in this folding,
except maybe at the very top

414
00:20:29,760 --> 00:20:31,710
and bottom.

415
00:20:31,710 --> 00:20:34,160
Except, we're only considering
half of the H nodes,

416
00:20:34,160 --> 00:20:37,620
so there's the other half
we have no idea about.

417
00:20:37,620 --> 00:20:41,080
So hopefully, nature
chose the parity right.

418
00:20:41,080 --> 00:20:44,409
Or, if you think about this is
in the triangular case, which

419
00:20:44,409 --> 00:20:45,950
I don't think anyone's
tried to adapt

420
00:20:45,950 --> 00:20:47,790
this algorithm to
the triangular grid,

421
00:20:47,790 --> 00:20:49,220
maybe you can get
no exposed node.

422
00:20:49,220 --> 00:20:51,530
So that would be pretty neat.

423
00:20:51,530 --> 00:20:54,127
Of course reality, it probably
does not fold on a lattice,

424
00:20:54,127 --> 00:20:55,210
but it's an approximation.

425
00:21:03,930 --> 00:21:10,470
Now, one of the observed
features in protein folding

426
00:21:10,470 --> 00:21:13,440
is they tend to fold
to the same shape.

427
00:21:13,440 --> 00:21:15,040
It's actually a hard
thing to measure,

428
00:21:15,040 --> 00:21:17,910
but that's the
general consensus,

429
00:21:17,910 --> 00:21:21,840
that at least most proteins
in a non diseased organism

430
00:21:21,840 --> 00:21:25,000
fold always to the
same shape, at least

431
00:21:25,000 --> 00:21:26,650
in the same environment.

432
00:21:26,650 --> 00:21:29,570
So if you want to model
that in the HP model,

433
00:21:29,570 --> 00:21:33,421
and you pretend that, really, we
are finding optimal solutions--

434
00:21:33,421 --> 00:21:34,920
because it's a
little hard to define

435
00:21:34,920 --> 00:21:36,859
what locally optimal means.

436
00:21:36,859 --> 00:21:38,650
But that would be an
interesting direction.

437
00:21:38,650 --> 00:21:40,390
It would be nice
to find proteins

438
00:21:40,390 --> 00:21:43,380
where locally optimal
equals globally optimal,

439
00:21:43,380 --> 00:21:46,060
but lacking definitions
of those terms,

440
00:21:46,060 --> 00:21:47,550
no one has tried to prove that.

441
00:21:47,550 --> 00:21:49,250
It could pick a good direction.

442
00:21:49,250 --> 00:21:52,530
But if you believe in
optimality, then at least

443
00:21:52,530 --> 00:21:56,135
you hope that the
optimal are unique.

444
00:21:56,135 --> 00:21:58,010
So you take a typical
example of the optimal,

445
00:21:58,010 --> 00:22:01,830
like if they're all blue,
then every folding is optimal.

446
00:22:01,830 --> 00:22:04,210
So there's exponentially
many of them.

447
00:22:04,210 --> 00:22:06,970
But if you have any hope
of getting unique folding,

448
00:22:06,970 --> 00:22:11,400
you'd really like
unique optimal foldings.

449
00:22:11,400 --> 00:22:12,605
The theorem is these exist.

450
00:22:15,330 --> 00:22:19,190
And the 2D squared
grid, they exist

451
00:22:19,190 --> 00:22:29,409
for all even n, for
closed chains-- mostly,

452
00:22:29,409 --> 00:22:31,450
we care about open chains,
but for closed chains,

453
00:22:31,450 --> 00:22:34,210
you could prove these
exist for all even n--

454
00:22:34,210 --> 00:22:42,617
and for open chains, they
exist for all doubly even n.

455
00:22:50,250 --> 00:22:53,680
This is a theorem by
a bunch of people.

456
00:22:53,680 --> 00:22:56,490
Oswin Aicholzer, David
Bremner, me, Hank Meijer,

457
00:22:56,490 --> 00:22:59,190
Vera Sacristan,
and Michael Soss,

458
00:22:59,190 --> 00:23:02,640
we talked about last time.

459
00:23:02,640 --> 00:23:04,580
So here they are
for closed chains.

460
00:23:04,580 --> 00:23:08,950
It's a pretty simple example,
basically alternating red blue,

461
00:23:08,950 --> 00:23:15,900
but at the end, you have two
blue nodes to make the corner.

462
00:23:15,900 --> 00:23:17,410
That is not the optimal folding.

463
00:23:17,410 --> 00:23:18,750
The optimal folding is this.

464
00:23:22,610 --> 00:23:25,710
So there's only
two exposed ends.

465
00:23:25,710 --> 00:23:28,370
There's one down here,
and one on the top.

466
00:23:28,370 --> 00:23:30,760
So there's no surprises that
it's optimal, a little less

467
00:23:30,760 --> 00:23:34,270
obvious that it's
uniquely optimal.

468
00:23:34,270 --> 00:23:35,310
But there it is.

469
00:23:35,310 --> 00:23:37,605
And this works for any even n.

470
00:23:37,605 --> 00:23:43,270
In general, it depends a little
bit on your the n mod four, so

471
00:23:43,270 --> 00:23:45,125
that's why we are drawing
two pictures here,

472
00:23:45,125 --> 00:23:49,330
to check that it works for
both congruences mod four.

473
00:23:49,330 --> 00:23:50,990
It has to be even
for closed chain

474
00:23:50,990 --> 00:23:53,510
and must be even, so
this is really for all m

475
00:23:53,510 --> 00:23:55,240
that there's closed
chains, there

476
00:23:55,240 --> 00:23:59,650
are unique-- there are
examples of HP chains

477
00:23:59,650 --> 00:24:02,320
that have unique
optimal foldings.

478
00:24:02,320 --> 00:24:05,280
Here, it's just to do you end
going horizontal or n going

479
00:24:05,280 --> 00:24:07,360
vertical?

480
00:24:07,360 --> 00:24:09,640
Not a big difference.

481
00:24:09,640 --> 00:24:16,970
I have a proof by picture that
this is the optimal folding.

482
00:24:16,970 --> 00:24:19,980
So remember there
are two-- we talk

483
00:24:19,980 --> 00:24:23,190
about H-H edges and H-H bonds.

484
00:24:23,190 --> 00:24:27,470
These are what we
call P-P edges.

485
00:24:27,470 --> 00:24:30,640
And I remember when writing this
paper, one of the co-authors,

486
00:24:30,640 --> 00:24:33,030
I won't say which one,
saying call me juvenile,

487
00:24:33,030 --> 00:24:35,840
but I don't want to have a paper
that talks about P-P edges.

488
00:24:35,840 --> 00:24:37,820
And I convinced him to keep it.

489
00:24:37,820 --> 00:24:40,820
So we have-- it's
really only obvious

490
00:24:40,820 --> 00:24:44,090
when you say it out loud,
which sadly I have to do here.

491
00:24:44,090 --> 00:24:46,720
But you have two of those
edges, all blue edges,

492
00:24:46,720 --> 00:24:48,930
we might also call them.

493
00:24:48,930 --> 00:24:53,120
And they really should
be on the boundary

494
00:24:53,120 --> 00:24:56,400
because you have to have
exposed ends on the boundary,

495
00:24:56,400 --> 00:24:59,040
on the bounding
box is what I mean.

496
00:24:59,040 --> 00:25:01,872
So if you put too many red
guys on the bounding box,

497
00:25:01,872 --> 00:25:03,330
you're going to
lose lots of bonds.

498
00:25:03,330 --> 00:25:06,150
So those two blue guys
should be on the boundary.

499
00:25:06,150 --> 00:25:07,900
Some two edges have
to be on the boundary,

500
00:25:07,900 --> 00:25:10,330
so it's best to have
the two blue ones.

501
00:25:10,330 --> 00:25:13,710
So we could show you in optimal
solutions, that is the case.

502
00:25:13,710 --> 00:25:15,890
Now, you decompose
into two chains

503
00:25:15,890 --> 00:25:17,060
that connect the two ends.

504
00:25:17,060 --> 00:25:19,480
There's connecting
this end to the left

505
00:25:19,480 --> 00:25:23,450
and connecting this end up and
around to the other blue node

506
00:25:23,450 --> 00:25:24,350
at the end.

507
00:25:24,350 --> 00:25:28,490
And if you check all of
the red notes on this side

508
00:25:28,490 --> 00:25:31,226
have even parity, let's say,
and all the red nodes on this

509
00:25:31,226 --> 00:25:33,660
die have odd parity
because it alternates.

510
00:25:33,660 --> 00:25:37,320
So there's no way
that you could have

511
00:25:37,320 --> 00:25:38,990
bonds within one
of those chains.

512
00:25:38,990 --> 00:25:40,795
That's not drawn here.

513
00:25:40,795 --> 00:25:42,920
The other thing is because
these two blue edges are

514
00:25:42,920 --> 00:25:47,986
on the boundary, you cannot
have a bond on the outside

515
00:25:47,986 --> 00:25:49,360
because there's
no hope for that.

516
00:25:49,360 --> 00:25:53,560
You can only have bonds on the
inside of that closed loop.

517
00:25:53,560 --> 00:25:57,480
Here, we're using that
it's a closed loop.

518
00:25:57,480 --> 00:26:00,000
So you have bonds on the inside,
and then you argue, well,

519
00:26:00,000 --> 00:26:02,980
because it's bipartite graph
and because it's planar,

520
00:26:02,980 --> 00:26:05,160
you can't have any
crossings in these bonds

521
00:26:05,160 --> 00:26:08,170
because you're embedding
this thing in the plane,

522
00:26:08,170 --> 00:26:11,570
really you basically
have to alternate.

523
00:26:11,570 --> 00:26:14,220
There's actually two ways
you could try to alternate.

524
00:26:14,220 --> 00:26:16,770
You start on this side or
if you start on this side.

525
00:26:16,770 --> 00:26:18,730
You try to realize
them geometrically.

526
00:26:18,730 --> 00:26:20,980
When you do this alternation,
you decompose your thing

527
00:26:20,980 --> 00:26:22,870
into squares.

528
00:26:22,870 --> 00:26:25,840
Only in one case does it work,
and you get this picture.

529
00:26:25,840 --> 00:26:27,480
And then once you
have the squares,

530
00:26:27,480 --> 00:26:29,692
there's a unique way
to glue them together.

531
00:26:29,692 --> 00:26:30,900
So that's how the proof goes.

532
00:26:30,900 --> 00:26:35,390
It's pretty easy once you
get the right limits in place

533
00:26:35,390 --> 00:26:36,730
for closed chains.

534
00:26:36,730 --> 00:26:40,110
Let me show you what
happens for open chains.

535
00:26:40,110 --> 00:26:44,930
Take the same example for open
chains, and it's almost unique.

536
00:26:44,930 --> 00:26:49,730
In the double even case,
like 16, it is unique.

537
00:26:49,730 --> 00:26:54,310
In the only singlely even,
so two mod four case,

538
00:26:54,310 --> 00:26:59,000
like 18, you have this issue.

539
00:26:59,000 --> 00:27:00,740
So the way we set
up is we do not

540
00:27:00,740 --> 00:27:05,560
have the blue blue at
the end to turn around.

541
00:27:05,560 --> 00:27:08,830
That is the best
we know how to do.

542
00:27:08,830 --> 00:27:13,160
And there are
these two-foldings,

543
00:27:13,160 --> 00:27:15,190
how you might achieve it.

544
00:27:15,190 --> 00:27:20,920
So here, we wrap around and get
a little triad of blues here.

545
00:27:20,920 --> 00:27:22,910
Because we have
a red n point, we

546
00:27:22,910 --> 00:27:25,260
can actually get
three bonds into it.

547
00:27:25,260 --> 00:27:27,595
Or, we can do the
intended folding.

548
00:27:27,595 --> 00:27:31,920
Now, it's a pretty small change,
so it's approximately unique,

549
00:27:31,920 --> 00:27:36,600
but that's the best we know for
open chains, only singly even

550
00:27:36,600 --> 00:27:37,170
length.

551
00:27:37,170 --> 00:27:39,650
Or for odd length, we also don't
know anything that's optimal.

552
00:27:39,650 --> 00:27:41,200
Of course there are things
that are almost optimal.

553
00:27:41,200 --> 00:27:42,490
You just add a little end.

554
00:27:42,490 --> 00:27:48,790
Won't make a big difference,
but interesting open problem.

555
00:27:48,790 --> 00:27:51,360
Also open, whether any
of these unique results--

556
00:27:51,360 --> 00:27:53,420
uniqueness results
extend to other grids,

557
00:27:53,420 --> 00:27:54,410
like triangular grid.

558
00:27:54,410 --> 00:27:57,034
That would be nice because here
we're really exploiting parity,

559
00:27:57,034 --> 00:27:58,330
and that's kind of cheating.

560
00:27:58,330 --> 00:28:00,580
Certainly, parity is not
such a big deal in real life,

561
00:28:00,580 --> 00:28:01,800
but I don't know.

562
00:28:01,800 --> 00:28:04,030
Could be.

563
00:28:04,030 --> 00:28:07,070
The approximation algorithm's
generalized to other grids,

564
00:28:07,070 --> 00:28:10,390
not necessarily this one,
but some constant factor

565
00:28:10,390 --> 00:28:13,180
approximation have
been obtained.

566
00:28:13,180 --> 00:28:14,470
Uniqueness is open.

567
00:28:14,470 --> 00:28:16,570
The really cool open
question here, I think,

568
00:28:16,570 --> 00:28:19,020
is protein design.

569
00:28:19,020 --> 00:28:21,390
HP model is simple enough
we can think about it.

570
00:28:21,390 --> 00:28:25,430
And while folding-- finding
the optimal folding of a given

571
00:28:25,430 --> 00:28:28,760
protein is NP-hard, if I
get to design the protein,

572
00:28:28,760 --> 00:28:32,170
and say, well, I'd really like
to fold into-- I don't know.

573
00:28:32,170 --> 00:28:36,040
The letter m-- how do I make a
protein fold into the letter m,

574
00:28:36,040 --> 00:28:37,800
at least approximately?

575
00:28:37,800 --> 00:28:41,080
So ideally, it should still
have a unique optimal folding,

576
00:28:41,080 --> 00:28:43,090
so we generalize
this result, but it

577
00:28:43,090 --> 00:28:45,260
can match any target shape.

578
00:28:45,260 --> 00:28:47,490
Here, we're matching
a diagonal line

579
00:28:47,490 --> 00:28:49,390
up to some constant
factor resolution.

580
00:28:49,390 --> 00:28:51,570
Can we match any shape up
to some constant factor

581
00:28:51,570 --> 00:28:52,120
resolution?

582
00:28:52,120 --> 00:28:53,734
I think the answer
should be yes.

583
00:28:53,734 --> 00:28:56,150
And then you should be able
to solve that polynomial time.

584
00:28:56,150 --> 00:28:58,350
In the same way that origami
design is a lot easier

585
00:28:58,350 --> 00:29:00,269
than origami
foldability, here, I

586
00:29:00,269 --> 00:29:02,060
think protein design
should be a lot easier

587
00:29:02,060 --> 00:29:04,960
than protein folding.

588
00:29:04,960 --> 00:29:06,460
But no one has worked on that.

589
00:29:06,460 --> 00:29:08,619
I think it's a
really cool problem.

590
00:29:08,619 --> 00:29:10,660
To bad I'm mentioning it
so late in the semester,

591
00:29:10,660 --> 00:29:12,535
but we can work on it
in the problem session.

592
00:29:15,800 --> 00:29:18,750
So ends my coverage
of protein folding.

593
00:29:18,750 --> 00:29:20,862
Obviously, there's tons
of work I'm not covering,

594
00:29:20,862 --> 00:29:22,820
but I'm focusing on the
very algorithmic stuff.

595
00:29:22,820 --> 00:29:24,990
Any questions before we go?

596
00:29:24,990 --> 00:29:25,778
Yeah?

597
00:29:25,778 --> 00:29:27,770
AUDIENCE: In these models
you have, are these

598
00:29:27,770 --> 00:29:31,320
supposed to equal
number of P and H,

599
00:29:31,320 --> 00:29:34,160
can you get in situations
where you're way

600
00:29:34,160 --> 00:29:37,342
overbeared with P
or H, and then it's

601
00:29:37,342 --> 00:29:39,700
going to be harder to
find an optimal folding.

602
00:29:39,700 --> 00:29:42,945
Or, is it not
realistic because these

603
00:29:42,945 --> 00:29:47,094
are such-- these are small
sets of the entire protein?

604
00:29:47,094 --> 00:29:47,760
PROFESSOR: Yeah.

605
00:29:47,760 --> 00:29:50,190
So the question
is whether-- Does

606
00:29:50,190 --> 00:29:53,180
it matter the ratio between the
number of H nodes and P nodes.

607
00:29:53,180 --> 00:29:57,724
I think the easy-- A sure
thing is I don't know.

608
00:29:57,724 --> 00:29:59,390
My guess is, in
reality, the number of H

609
00:29:59,390 --> 00:30:02,310
and P's are within a constant
factor of each other,

610
00:30:02,310 --> 00:30:03,170
for some constant.

611
00:30:03,170 --> 00:30:05,020
I don't think the
constant matters too much.

612
00:30:05,020 --> 00:30:06,853
Although if there's an
overwhelming majority

613
00:30:06,853 --> 00:30:09,682
of H's or P's, the problem
might become easier.

614
00:30:09,682 --> 00:30:11,390
My guess also is in
the NP-hardness proof

615
00:30:11,390 --> 00:30:13,800
although it's been a
while since I read it.

616
00:30:13,800 --> 00:30:16,967
There's probably a
constant factor again.

617
00:30:16,967 --> 00:30:19,050
But maybe you could show
if there's-- for example,

618
00:30:19,050 --> 00:30:21,356
if there's a constant
number of h nodes,

619
00:30:21,356 --> 00:30:23,230
I'll bet you can solve
it in polynomial time.

620
00:30:23,230 --> 00:30:25,430
That's maybe-- That's
a new open problem.

621
00:30:25,430 --> 00:30:26,570
I should write it down.

622
00:30:26,570 --> 00:30:27,710
Good question.

623
00:30:27,710 --> 00:30:31,686
AUDIENCE: If there is a
wrong factor between them,

624
00:30:31,686 --> 00:30:34,171
wouldn't we see
that at the model

625
00:30:34,171 --> 00:30:37,020
because it just wouldn't work?

626
00:30:37,020 --> 00:30:40,420
PROFESSOR: No, this model makes
sense for any number of reds

627
00:30:40,420 --> 00:30:44,520
and H's and P's, but uniqueness
is probably hard to obtain

628
00:30:44,520 --> 00:30:47,650
if you have a small number
of H's or if you have a huge

629
00:30:47,650 --> 00:30:48,590
number of H's.

630
00:30:48,590 --> 00:30:50,596
If you don't have enough--

631
00:30:50,596 --> 00:30:54,670
AUDIENCE: What if you have
H's on the order of the square

632
00:30:54,670 --> 00:30:57,892
of the P's, then there
are at least two H's?

633
00:30:57,892 --> 00:30:59,760
PROFESSOR: All right.

634
00:30:59,760 --> 00:31:01,940
AUDIENCE: Kind of
makes sense, right?

635
00:31:01,940 --> 00:31:04,690
PROFESSOR: Yeah,
actually we have

636
00:31:04,690 --> 00:31:07,960
a theorem along those lines,
which is not published

637
00:31:07,960 --> 00:31:10,700
so that's why I'm
not talking about it.

638
00:31:10,700 --> 00:31:13,700
You can achieve the number
of H's as square the number

639
00:31:13,700 --> 00:31:15,010
of P's.

640
00:31:15,010 --> 00:31:17,260
I imagine that's the limit--
and still get uniqueness.

641
00:31:17,260 --> 00:31:18,450
I imagine that's the limit.

642
00:31:18,450 --> 00:31:19,949
We really ought to
write that result

643
00:31:19,949 --> 00:31:24,399
that it's from seven years
ago, so unlikely at this point.

644
00:31:24,399 --> 00:31:25,065
Other questions?

645
00:31:31,120 --> 00:31:43,780
So we go onto
interlocked 3D chains,

646
00:31:43,780 --> 00:31:47,060
which is related in that
it's in 3D and its chains.

647
00:31:47,060 --> 00:31:50,280
So related to the
mechanical protein folding

648
00:31:50,280 --> 00:31:52,810
we did last class,
except here, mostly,

649
00:31:52,810 --> 00:31:54,810
I'm going to think about
universal joints again,

650
00:31:54,810 --> 00:31:56,167
so I can do whatever I want.

651
00:31:56,167 --> 00:31:57,750
I don't have to hold
the angles fixed.

652
00:31:57,750 --> 00:31:59,208
At the end, I'll
mention how things

653
00:31:59,208 --> 00:32:02,515
change a little bit for when
you have fixed angle chains.

654
00:32:05,340 --> 00:32:07,180
But universals more fun.

655
00:32:07,180 --> 00:32:09,540
So the motivation
here is we have

656
00:32:09,540 --> 00:32:12,870
our good friend the
knitting needles example,

657
00:32:12,870 --> 00:32:14,560
or maybe our mortal
enemy, depending

658
00:32:14,560 --> 00:32:16,130
on how you like
to think of that.

659
00:32:16,130 --> 00:32:20,040
One, two, three, four, five.

660
00:32:20,040 --> 00:32:22,310
And remember-- never
mind the edge lengths.

661
00:32:22,310 --> 00:32:24,330
I'm not going to worry
about edge links here.

662
00:32:24,330 --> 00:32:26,865
With five bars, you
can lock the chain.

663
00:32:30,760 --> 00:32:33,630
So an interesting
question is what

664
00:32:33,630 --> 00:32:39,330
if I had less than five
chains, say length 4 chains,

665
00:32:39,330 --> 00:32:42,310
but I had more than one of them.

666
00:32:42,310 --> 00:32:45,050
Now certainly, each
chain will be unlocked.

667
00:32:45,050 --> 00:32:49,530
That's a known result by
Cantrell and Johnson '96.

668
00:32:49,530 --> 00:32:50,610
'96, '98?

669
00:32:50,610 --> 00:32:52,350
'96.

670
00:32:52,350 --> 00:32:57,230
But what if I have two four
chains, can they interlock?

671
00:32:57,230 --> 00:32:59,790
The answer turns out to be yes.

672
00:32:59,790 --> 00:33:02,280
The original motivation
for this proble,--

673
00:33:02,280 --> 00:33:07,560
it's a neat question by itself,
but the original motivation is

674
00:33:07,560 --> 00:33:10,750
something called
Lubiw's problem.

675
00:33:10,750 --> 00:33:15,800
Many of you met Anna Lubiw,
she's one of my Ph.D. Advisors.

676
00:33:15,800 --> 00:33:19,360
Last year, she was
here on sabbatical.

677
00:33:19,360 --> 00:33:21,420
And the question
is the following--

678
00:33:21,420 --> 00:33:25,100
so she posed this back
in 2000, a long time ago.

679
00:33:25,100 --> 00:33:29,000
What is the minimum
number of cuts

680
00:33:29,000 --> 00:33:34,000
you have to make in a 3D
chain in order to unlock it?

681
00:33:49,690 --> 00:33:55,720
I say unlocked meaning we
know they are locked chains.

682
00:33:55,720 --> 00:33:58,420
Suppose I want to make the
configuration space connected,

683
00:33:58,420 --> 00:34:00,610
but I allow cheating and
I allow cutting vertices.

684
00:34:00,610 --> 00:34:02,240
How many do I have to cut?

685
00:34:02,240 --> 00:34:06,070
And say the chain has n bars.

686
00:34:06,070 --> 00:34:08,909
Then, as a function of n,
how many cuts do I need?

687
00:34:12,730 --> 00:34:15,630
And the answer at this
point, in the worst case,

688
00:34:15,630 --> 00:34:19,870
we know is between about
n over 2 and n over 4.

689
00:34:19,870 --> 00:34:30,190
So it's at least 4
over n minus 1 over 4.

690
00:34:30,190 --> 00:34:37,210
And it's at most 4
n minus 3 over 2.

691
00:34:37,210 --> 00:34:39,350
And I'll prove to you
both of those results.

692
00:34:39,350 --> 00:34:41,877
So we know roughly its theta n.

693
00:34:41,877 --> 00:34:43,460
It's in a constant
factor of n, but we

694
00:34:43,460 --> 00:34:45,480
don't know exactly
what the constant is,

695
00:34:45,480 --> 00:34:47,300
still a neat problem.

696
00:34:47,300 --> 00:34:50,530
Originally this motivation
here is, well, maybe-- there's

697
00:34:50,530 --> 00:34:52,969
some theories that
in real proteins,

698
00:34:52,969 --> 00:34:56,050
maybe they disconnect their
bonds for a little while

699
00:34:56,050 --> 00:34:59,085
so they can pass through
and make things cross.

700
00:34:59,085 --> 00:35:00,460
How many disconnections
would you

701
00:35:00,460 --> 00:35:03,370
need because that might
require a lot of energy?

702
00:35:03,370 --> 00:35:07,600
The answer is a lot if
you have a long protein.

703
00:35:07,600 --> 00:35:09,580
in the worst case of course.

704
00:35:09,580 --> 00:35:12,440
Real proteins have unit edge
lengths almost equal angles,

705
00:35:12,440 --> 00:35:14,784
so maybe the answers
change in that case,

706
00:35:14,784 --> 00:35:16,700
but given that we don't
even know whether it's

707
00:35:16,700 --> 00:35:19,070
a locked chain, in
those situations,

708
00:35:19,070 --> 00:35:20,940
it's pretty hard
for us to answer

709
00:35:20,940 --> 00:35:24,090
something more
complicated like this.

710
00:35:24,090 --> 00:35:26,240
So let me start with
the lower bound.

711
00:35:26,240 --> 00:35:29,245
You need to make at least
n over 4 cuts almost.

712
00:35:31,770 --> 00:35:34,310
Any ideas how to do that
before I show you the answer?

713
00:35:38,680 --> 00:35:41,701
Given the one tool we have,
which is the knitting needles.

714
00:35:41,701 --> 00:35:44,674
AUDIENCE: Yeah, I mean
if you [INAUDIBLE].

715
00:35:44,674 --> 00:35:46,215
PROFESSOR: Just
connect a whole bunch

716
00:35:46,215 --> 00:35:47,100
of knitting needles together.

717
00:35:47,100 --> 00:35:47,770
Yeah.

718
00:35:47,770 --> 00:35:49,497
Now, knitting needles
has five edges,

719
00:35:49,497 --> 00:35:51,080
so that would naturally
give n over 5.

720
00:35:51,080 --> 00:35:54,224
But if you're a little bit
clever, you can share edges.

721
00:35:54,224 --> 00:35:56,390
If you share the long edges
from one knitting needle

722
00:35:56,390 --> 00:36:02,780
to the next, and you get about
4 bars per knitting needle,

723
00:36:02,780 --> 00:36:05,740
except the very first bar which
have to count extra, and so

724
00:36:05,740 --> 00:36:08,480
that's the n minus
1 divided by 4.

725
00:36:08,480 --> 00:36:13,540
Now if you don't cut
one of the vertices--

726
00:36:13,540 --> 00:36:16,424
and here, we're
considering vertex cuts--

727
00:36:16,424 --> 00:36:17,840
if you don't cut
one of those four

728
00:36:17,840 --> 00:36:19,450
vertices of that
knitting needle,

729
00:36:19,450 --> 00:36:20,820
the whole thing will be
locked because that's

730
00:36:20,820 --> 00:36:21,720
a knitting needle.

731
00:36:21,720 --> 00:36:24,220
So you have to cut one of those
four, and one of those four,

732
00:36:24,220 --> 00:36:28,120
and one of those four just for
the pieces to not be locked.

733
00:36:28,120 --> 00:36:31,070
The trouble is-- so the natural
algorithm, if I give you

734
00:36:31,070 --> 00:36:35,940
some big complicated chain,
is cut every fourth vertex.

735
00:36:35,940 --> 00:36:39,040
That we know does not work.

736
00:36:39,040 --> 00:36:44,350
I think I have to
wait to see that one.

737
00:36:44,350 --> 00:36:46,410
Let me tell you what's
known about interlocking,

738
00:36:46,410 --> 00:36:48,570
and then we'll get back
to Lubiw's problem.

739
00:36:48,570 --> 00:36:51,100
But at least we've
proved this part.

740
00:36:51,100 --> 00:36:53,670
And next thing I want to
do is prove this part.

741
00:36:53,670 --> 00:36:57,690
And to show you how
we might do that, I

742
00:36:57,690 --> 00:36:58,830
think I need a whole board.

743
00:37:14,140 --> 00:37:21,742
We're going to think about
an open chain of length 2--

744
00:37:21,742 --> 00:37:24,210
let me make a little
more room-- 2,

745
00:37:24,210 --> 00:37:41,220
3, 4 interlocking with-- start
with an open chain of length 2,

746
00:37:41,220 --> 00:37:42,620
3, or 4.

747
00:37:42,620 --> 00:37:45,520
There's going to be other
results, but it's a good start.

748
00:37:48,310 --> 00:37:55,127
So for 2, 2 chains, I
claim are always separable.

749
00:37:55,127 --> 00:37:56,710
And I'm going to put
a little asterisk

750
00:37:56,710 --> 00:37:58,650
to mean another thing is true.

751
00:37:58,650 --> 00:38:01,030
A 2 chain and a 3 chain
is always separable.

752
00:38:01,030 --> 00:38:03,890
A 2 chain and a 4 chain
is always separable.

753
00:38:03,890 --> 00:38:05,520
This matrix is symmetric.

754
00:38:11,136 --> 00:38:15,080
A 3 chain versus a 3
chain also is separable.

755
00:38:18,180 --> 00:38:19,805
But everything else,
you can interlock.

756
00:38:24,610 --> 00:38:26,920
I guess I haven't
defined interlock.

757
00:38:26,920 --> 00:38:28,660
Be a good idea.

758
00:38:28,660 --> 00:38:31,220
I call a bunch of-- a
collection of chains

759
00:38:31,220 --> 00:38:34,276
separable if they can all fly to
infinity away from each other.

760
00:38:34,276 --> 00:38:35,650
So the distances
between them can

761
00:38:35,650 --> 00:38:37,858
get arbitrarily large,
otherwise they're interlocked.

762
00:38:41,000 --> 00:38:43,000
And usually, that will
mean that no pair of them

763
00:38:43,000 --> 00:38:44,330
can fly away from each other.

764
00:38:44,330 --> 00:38:45,350
Here, we're just
thinking about two

765
00:38:45,350 --> 00:38:47,391
at once, so it's either
separable or interlocked.

766
00:38:50,120 --> 00:38:53,230
So this is a worry
if you're thinking

767
00:38:53,230 --> 00:38:54,700
about cutting this apart.

768
00:38:54,700 --> 00:38:57,575
You cut every fourth
bar-- fourth vertex,

769
00:38:57,575 --> 00:38:59,380
if I get the counting
right, I think

770
00:38:59,380 --> 00:39:02,720
that means you have length
4 chains connecting them.

771
00:39:02,720 --> 00:39:05,170
Now, we know length 4
chains can't be locked,

772
00:39:05,170 --> 00:39:06,880
but they could be interlocked.

773
00:39:06,880 --> 00:39:09,170
And so you could
actually build an example

774
00:39:09,170 --> 00:39:12,420
where if you do the simple
pattern of every fourth cut,

775
00:39:12,420 --> 00:39:13,895
you will get interlocking.

776
00:39:13,895 --> 00:39:17,260
It's still open whether you
could be clever and still

777
00:39:17,260 --> 00:39:19,800
only make about n over 4
cuts and get everything

778
00:39:19,800 --> 00:39:22,740
to separate, but to solve
that, you'd have to get around

779
00:39:22,740 --> 00:39:26,330
this boundary, this annoying
situation, which we don't fully

780
00:39:26,330 --> 00:39:27,150
have characterized.

781
00:39:27,150 --> 00:39:29,420
We just have examples.

782
00:39:29,420 --> 00:39:31,300
And we also thought
about-- and we'll

783
00:39:31,300 --> 00:39:34,680
get to this later--
if you try interlock

784
00:39:34,680 --> 00:39:36,650
an open chain with
a closed chain.

785
00:39:36,650 --> 00:39:38,350
With two closed
chains, it's trivial.

786
00:39:38,350 --> 00:39:42,800
You can just make a
knot, or make a link.

787
00:39:42,800 --> 00:39:44,990
I guess is the proper term.

788
00:39:44,990 --> 00:39:48,915
But an open versus a closed is
more interesting, especially

789
00:39:48,915 --> 00:39:52,710
if that closed guy
is not itself a knot.

790
00:39:52,710 --> 00:39:56,640
I think up to length 5.

791
00:39:56,640 --> 00:39:58,930
Yeah, I think the
smallest, locked,

792
00:39:58,930 --> 00:40:01,350
closed chain is of length 6.

793
00:40:01,350 --> 00:40:06,050
Never showed it to you,
but that is a known fact.

794
00:40:06,050 --> 00:40:09,290
So up to 5, we know this
thing is not by itself locked,

795
00:40:09,290 --> 00:40:12,150
so it's not really cheating
compared to an open chain.

796
00:40:12,150 --> 00:40:16,750
Then, the answer is separable,
separable, separable.

797
00:40:16,750 --> 00:40:19,920
It's really hard to lock
things with a 2 chain.

798
00:40:19,920 --> 00:40:23,162
2 chains, we call hairpins.

799
00:40:23,162 --> 00:40:24,170
That's a 2 chain.

800
00:40:32,649 --> 00:40:33,940
Everything else is interlocked.

801
00:40:38,870 --> 00:40:41,595
Basically, the same results
hold whether one of your chains

802
00:40:41,595 --> 00:40:47,272
is opened or closed because
notice there's a shift here.

803
00:40:47,272 --> 00:40:48,980
The smallest closed
chains have length 3,

804
00:40:48,980 --> 00:40:50,850
and the smallest open chain
is really of length 2.

805
00:40:50,850 --> 00:40:52,510
I guess you could
consider single bars,

806
00:40:52,510 --> 00:40:54,170
but they don't usually do much.

807
00:40:54,170 --> 00:40:57,250
You're can just slide them
out parallel to themselves.

808
00:40:57,250 --> 00:40:58,770
So there's a shift
in the indices

809
00:40:58,770 --> 00:41:01,785
here, and that's why there's
a shift in the matrix.

810
00:41:01,785 --> 00:41:02,410
Sounds ominous.

811
00:41:02,410 --> 00:41:06,560
A shift in the matrix,
what will we do?

812
00:41:06,560 --> 00:41:08,410
But they're basically the same.

813
00:41:08,410 --> 00:41:11,430
So I'm going to show most
of these results to you.

814
00:41:11,430 --> 00:41:15,850
The stars mean that even if
I give you-- get this right--

815
00:41:15,850 --> 00:41:18,930
even if I give you a whole
bunch more hairpins-- so

816
00:41:18,930 --> 00:41:20,860
for this result, it's
the most interesting.

817
00:41:20,860 --> 00:41:25,640
I take a 3 chain, a 3 chain,
and then a millionaire hairpins,

818
00:41:25,640 --> 00:41:27,240
a million 2 chains.

819
00:41:27,240 --> 00:41:29,462
Still, they can separate.

820
00:41:29,462 --> 00:41:30,920
And all the results
with stars, you

821
00:41:30,920 --> 00:41:32,336
can add arbitrarily
many hairpins.

822
00:41:35,950 --> 00:41:38,000
These two are the most
interesting, a 4 chain

823
00:41:38,000 --> 00:41:39,860
with any number
of hairpins, a 2,

824
00:41:39,860 --> 00:41:42,030
3 chain with any
number of hairpins.

825
00:41:42,030 --> 00:41:44,494
These results do
not hold or we don't

826
00:41:44,494 --> 00:41:45,660
know how to prove it anyway.

827
00:41:48,814 --> 00:41:49,605
That's the summary.

828
00:41:52,560 --> 00:41:56,950
There's one question that's
not answered by this table

829
00:41:56,950 --> 00:41:59,420
because it's how low can
you go and get interlocking.

830
00:41:59,420 --> 00:42:02,300
This is saying, well, once you
get to a 3 chain and a 4 chain,

831
00:42:02,300 --> 00:42:03,050
you can interlock.

832
00:42:03,050 --> 00:42:04,580
It's the minimum.

833
00:42:04,580 --> 00:42:09,290
But what if you have a bunch
of 2 chains or even just one?

834
00:42:09,290 --> 00:42:12,610
Can I interlock it with
some chain of some length?

835
00:42:12,610 --> 00:42:14,890
Let's say, I mean up to
4 is the most interesting

836
00:42:14,890 --> 00:42:17,264
because once you get to 5,
you can have knitting needles.

837
00:42:17,264 --> 00:42:21,930
But suppose I take longer
open chains down here,

838
00:42:21,930 --> 00:42:24,180
but I require that they're
not themselves locked.

839
00:42:24,180 --> 00:42:28,070
Can I interlock
it with a hairpin?

840
00:42:28,070 --> 00:42:31,555
It has practical
applications to hair I guess.

841
00:42:34,090 --> 00:42:36,410
And the answer is yes.

842
00:42:36,410 --> 00:42:41,290
So this first example
is a 16 chain.

843
00:42:41,290 --> 00:42:44,581
That's the gray part.

844
00:42:44,581 --> 00:42:45,080
Wow!

845
00:42:45,080 --> 00:42:46,740
It's hard to even
follow the ends.

846
00:42:46,740 --> 00:42:47,860
Here's one end.

847
00:42:47,860 --> 00:42:51,690
And here's another end
and 16 bars in the middle.

848
00:42:51,690 --> 00:42:54,740
And then the red
thing is a 2 chain,

849
00:42:54,740 --> 00:42:59,210
and this is a model of
that thing, physical model.

850
00:42:59,210 --> 00:43:00,830
And it's locked.

851
00:43:00,830 --> 00:43:02,900
Sorry, it's interlocked,
I should say.

852
00:43:02,900 --> 00:43:05,640
I'm pretty sure you take
either of the chains,

853
00:43:05,640 --> 00:43:07,410
obviously with the
hairpin, but also

854
00:43:07,410 --> 00:43:09,980
for the really
complicated 16 chain,

855
00:43:09,980 --> 00:43:11,130
it will unfold by itself.

856
00:43:11,130 --> 00:43:14,310
But with a hairpin, it is
held open at that angle

857
00:43:14,310 --> 00:43:16,850
and can't unfold.

858
00:43:19,710 --> 00:43:25,820
The best bound so far is an
11 chain versus a 2 chain.

859
00:43:25,820 --> 00:43:27,775
It's conjectured to be optimal.

860
00:43:27,775 --> 00:43:29,430
There's various
reasons to believe

861
00:43:29,430 --> 00:43:32,540
at least this interlocking
trick where you-- basically,

862
00:43:32,540 --> 00:43:37,350
all the hairpin can do is
hold an angle open or try to.

863
00:43:37,350 --> 00:43:41,510
And this seems to be the
smallest way to interlock it

864
00:43:41,510 --> 00:43:44,990
up there and the smallest way
to interlock it on either side.

865
00:43:44,990 --> 00:43:46,110
And these are zooms.

866
00:43:46,110 --> 00:43:47,490
This is of one of these corners.

867
00:43:47,490 --> 00:43:48,814
This is of the top part.

868
00:43:48,814 --> 00:43:50,230
So those, by
themselves, if you're

869
00:43:50,230 --> 00:43:52,980
trying to do these
three interlock tricks,

870
00:43:52,980 --> 00:43:55,610
these are minimal, we think.

871
00:43:55,610 --> 00:43:58,580
But as an assembly, it's
conjectured to be optimal,

872
00:43:58,580 --> 00:44:00,080
but we don't know
how to prove that.

873
00:44:00,080 --> 00:44:01,746
But at least these
guys are interlocked,

874
00:44:01,746 --> 00:44:03,670
so 11 chain with a 2 chain.

875
00:44:08,700 --> 00:44:12,030
So those don't fit on the
table because the table's not

876
00:44:12,030 --> 00:44:13,650
large enough.

877
00:44:13,650 --> 00:44:18,450
But let me start
with this result.

878
00:44:18,450 --> 00:44:22,790
So the easiest
thing to do is show

879
00:44:22,790 --> 00:44:33,150
that if I have any finite
set of hairpins, 2 chains,

880
00:44:33,150 --> 00:44:34,195
they never interlock.

881
00:44:38,494 --> 00:44:40,660
This is actually an old
result although it was never

882
00:44:40,660 --> 00:44:45,430
phrased this way until
earlier this decade,

883
00:44:45,430 --> 00:44:55,030
I guess, by de Brujin, a famous
Dutch mathematician from 1954.

884
00:44:55,030 --> 00:44:58,360
He proved essentially this.

885
00:44:58,360 --> 00:45:00,500
He proved a simpler
version which was then

886
00:45:00,500 --> 00:45:04,580
generalized in 1984 by
Robert Dawson in Canada.

887
00:45:04,580 --> 00:45:08,720
And they proved
something more general.

888
00:45:08,720 --> 00:45:13,250
They said, well-- I will say
a hairpin is star-shaped,

889
00:45:13,250 --> 00:45:17,550
meaning if you were right here,
here's your eye at this point,

890
00:45:17,550 --> 00:45:19,690
you can see the entire
interior of the shape

891
00:45:19,690 --> 00:45:21,270
without leaving the shape.

892
00:45:21,270 --> 00:45:24,120
So that's a star-shaped figure.

893
00:45:24,120 --> 00:45:26,980
And these guys proved that
for star-shaped figures,

894
00:45:26,980 --> 00:45:28,980
there's a way to
always separate them

895
00:45:28,980 --> 00:45:31,400
without even changing
any of the angles.

896
00:45:31,400 --> 00:45:33,490
There's no folding involved.

897
00:45:33,490 --> 00:45:35,490
You could think
of rigid 2 chains.

898
00:45:35,490 --> 00:45:37,760
They can all fly
apart from each other.

899
00:45:37,760 --> 00:45:39,400
How do you do it?

900
00:45:39,400 --> 00:45:40,610
Some fun geometry.

901
00:45:40,610 --> 00:45:47,982
You say, I have my favorite
0, 0, 0 in three space.

902
00:45:47,982 --> 00:45:49,940
I just want to scale
everything from the point,

903
00:45:49,940 --> 00:45:52,540
make space bigger.

904
00:45:52,540 --> 00:45:55,690
What will happen to
these poor little chains?

905
00:45:55,690 --> 00:45:56,944
They get bigger.

906
00:45:56,944 --> 00:45:57,735
They'll get longer.

907
00:46:00,620 --> 00:46:04,510
As they get longer,
just cut them off again.

908
00:46:04,510 --> 00:46:06,340
But the result is
that these points

909
00:46:06,340 --> 00:46:08,690
will fly away from each other.

910
00:46:08,690 --> 00:46:10,974
And so all of these two
chains are going to separate.

911
00:46:10,974 --> 00:46:13,390
The lengths get longer, but I
can always cut them shorter.

912
00:46:13,390 --> 00:46:15,215
And if I scale
space, obviously if I

913
00:46:15,215 --> 00:46:16,756
didn't have
self-intersection before,

914
00:46:16,756 --> 00:46:18,630
I still won't have
self-intersection.

915
00:46:18,630 --> 00:46:19,930
So it's a fun motion.

916
00:46:19,930 --> 00:46:22,310
We tried to draw it in the book.

917
00:46:22,310 --> 00:46:24,440
It's a little hard to draw.

918
00:46:24,440 --> 00:46:26,930
So we have the red, green,
and, blue-- hopefully,

919
00:46:26,930 --> 00:46:31,330
you're not colorblind--
hairpins hanging out.

920
00:46:31,330 --> 00:46:33,230
They're not hitting each other.

921
00:46:33,230 --> 00:46:34,360
Here's the origin.

922
00:46:34,360 --> 00:46:37,370
We scale everything, so the blue
guy goes to the dashed blue.

923
00:46:37,370 --> 00:46:39,390
Green goes to the dashed green.

924
00:46:39,390 --> 00:46:40,790
It looks weird.

925
00:46:40,790 --> 00:46:42,410
It's all over the place.

926
00:46:42,410 --> 00:46:44,160
And you're stretching
for this point,

927
00:46:44,160 --> 00:46:47,230
so as you go farther
away, it's bigger.

928
00:46:47,230 --> 00:46:49,690
Red guy expands to
the dashed red guy.

929
00:46:49,690 --> 00:46:52,372
What's not shown here is
then you clip the red things.

930
00:46:52,372 --> 00:46:53,830
Because everything
gets longer, you

931
00:46:53,830 --> 00:46:55,580
can always clip to
the original lengths.

932
00:46:55,580 --> 00:46:57,510
And that's a motion
of these chains.

933
00:46:57,510 --> 00:46:58,760
You're not changing the angle.

934
00:46:58,760 --> 00:47:00,930
Scaling preserves angles.

935
00:47:00,930 --> 00:47:03,830
And everything flies
way for each other.

936
00:47:03,830 --> 00:47:07,060
So a very nice trick
from a long time ago.

937
00:47:10,760 --> 00:47:14,640
Once you get to 3 chains,
this doesn't work.

938
00:47:14,640 --> 00:47:19,760
And so if we want to
prove this result, which

939
00:47:19,760 --> 00:47:23,850
would subsume these guys, we
need to do a little more work.

940
00:47:23,850 --> 00:47:26,520
Although a very
similar trick works.

941
00:47:26,520 --> 00:47:34,625
So let me show you why two
open 3 chains separate.

942
00:47:39,330 --> 00:47:44,160
Even if you add arbitrarily
many extra hairpins-- that's

943
00:47:44,160 --> 00:47:47,640
the star-- Two 3
chains always separate.

944
00:47:57,310 --> 00:47:59,270
I guess suppose we're
in a generic situation.

945
00:47:59,270 --> 00:48:01,210
Nothing aligns that shouldn't.

946
00:48:01,210 --> 00:48:05,230
I have one 3 chain, or
something like this.

947
00:48:05,230 --> 00:48:06,592
And I have another one.

948
00:48:06,592 --> 00:48:07,675
Let's make it interesting.

949
00:48:12,670 --> 00:48:14,050
Is that possible?

950
00:48:14,050 --> 00:48:15,020
Maybe.

951
00:48:15,020 --> 00:48:16,130
Something like that.

952
00:48:16,130 --> 00:48:18,886
Probably you want these
guys to be pretty long.

953
00:48:18,886 --> 00:48:21,200
Do some crazy whatever.

954
00:48:21,200 --> 00:48:25,680
That is two 3 chains
if I drew it right.

955
00:48:25,680 --> 00:48:27,780
And the claim is you can
always separate them.

956
00:48:37,900 --> 00:48:40,070
So here's the deal.

957
00:48:40,070 --> 00:48:42,350
I have this middle bar.

958
00:48:42,350 --> 00:48:46,290
There's a whole family of planes
that are parallel to that bar,

959
00:48:46,290 --> 00:48:47,690
in fact, a lot of planes.

960
00:48:47,690 --> 00:48:48,280
There's what?

961
00:48:48,280 --> 00:48:49,550
Three dimensions of planes.

962
00:48:49,550 --> 00:48:51,300
There's a two dimensional
family of planes

963
00:48:51,300 --> 00:48:52,780
that are parallel to that bar.

964
00:48:52,780 --> 00:48:55,280
There's a two dimensional family
of planes that are parallel

965
00:48:55,280 --> 00:48:57,080
to that bar, meaning
if you extend it,

966
00:48:57,080 --> 00:49:01,830
it doesn't hit the plane
or lies in the plane.

967
00:49:01,830 --> 00:49:05,422
So there's actually a plane
that is parallel to both bars.

968
00:49:05,422 --> 00:49:07,380
In fact, there's a whole
one dimensional family

969
00:49:07,380 --> 00:49:09,110
of that, if I
figured it out right.

970
00:49:09,110 --> 00:49:12,870
So you can actually find a
plane that bisects the two bars.

971
00:49:12,870 --> 00:49:16,140
So there's one on one side,
one on the other side.

972
00:49:16,140 --> 00:49:20,000
And even if the bars extended,
it would not hit the plane.

973
00:49:20,000 --> 00:49:23,700
So there's some plane
in the middle here.

974
00:49:23,700 --> 00:49:27,272
Really hard to draw in 3D.

975
00:49:27,272 --> 00:49:28,730
Actually, I think
I have a picture.

976
00:49:28,730 --> 00:49:31,990
It might be a little
easier to see.

977
00:49:31,990 --> 00:49:33,370
So it looks something like this.

978
00:49:33,370 --> 00:49:33,930
You have to believe me.

979
00:49:33,930 --> 00:49:34,880
This is general case.

980
00:49:34,880 --> 00:49:37,210
You have the red chain
and the blue chain.

981
00:49:37,210 --> 00:49:39,590
The middle bars are on
opposite sides of the plane,

982
00:49:39,590 --> 00:49:41,660
and they're parallel
to the plane.

983
00:49:41,660 --> 00:49:44,890
Now what we do is
scale, but we only

984
00:49:44,890 --> 00:49:47,991
scale in z, so
non-uniform scaling.

985
00:49:47,991 --> 00:49:49,240
This will not preserve angles.

986
00:49:49,240 --> 00:49:51,060
Here, we're going to fold.

987
00:49:51,060 --> 00:49:53,440
So the result, we scale
from z equals zero,

988
00:49:53,440 --> 00:49:55,110
so these guys get pushed down.

989
00:49:55,110 --> 00:49:56,730
Those guys get pushed up.

990
00:49:56,730 --> 00:50:00,340
Scaling in z, again, does not
cause any self-intersection.

991
00:50:00,340 --> 00:50:02,180
And because we made
these guys parallel,

992
00:50:02,180 --> 00:50:04,390
these edge lengths
will be preserved.

993
00:50:04,390 --> 00:50:05,850
That's the key.

994
00:50:05,850 --> 00:50:07,410
The end bars will
get longer, but I

995
00:50:07,410 --> 00:50:08,660
don't care if they get longer.

996
00:50:08,660 --> 00:50:10,817
I can always cut them shorter.

997
00:50:10,817 --> 00:50:11,650
So that's the trick.

998
00:50:13,869 --> 00:50:15,410
Yeah, they'll always
get longer, even

999
00:50:15,410 --> 00:50:17,810
if they don't go
to the other side.

1000
00:50:17,810 --> 00:50:19,840
So that's how you can
just pull them apart.

1001
00:50:19,840 --> 00:50:21,760
And also, if there
are 2 chains in here,

1002
00:50:21,760 --> 00:50:23,820
they will also just fly
away from each other.

1003
00:50:23,820 --> 00:50:25,020
So eventually,
everything will get

1004
00:50:25,020 --> 00:50:27,436
far away unless they happen
to have the same z-coordinate,

1005
00:50:27,436 --> 00:50:29,330
but then it's degenerate.

1006
00:50:29,330 --> 00:50:30,590
Just perturbed a little bit.

1007
00:50:30,590 --> 00:50:31,090
Jason?

1008
00:50:31,090 --> 00:50:34,278
AUDIENCE: Is this the
case where the two

1009
00:50:34,278 --> 00:50:38,260
bars are coplanar and parallel?

1010
00:50:38,260 --> 00:50:39,710
PROFESSOR: Yeah.

1011
00:50:39,710 --> 00:50:42,150
I said at the beginning
generic, so in particular,

1012
00:50:42,150 --> 00:50:44,977
I want to assume that these
guys are not coplanar,

1013
00:50:44,977 --> 00:50:47,060
otherwise, you could just
perturb it a little bit.

1014
00:50:47,060 --> 00:50:50,405
As long as they're not touching,
you can wiggle everybody.

1015
00:50:50,405 --> 00:50:52,920
Good point.

1016
00:50:52,920 --> 00:50:54,500
That's two 3 chains.

1017
00:50:54,500 --> 00:50:56,360
There's one more result
in the separability.

1018
00:50:56,360 --> 00:50:59,930
We've done all of these
separability results here.

1019
00:50:59,930 --> 00:51:03,450
The next one would be this,
which is the same as this.

1020
00:51:03,450 --> 00:51:07,480
So here we have a 4
chain versus a 2 chain.

1021
00:51:07,480 --> 00:51:09,470
Basically, the same trick works.

1022
00:51:09,470 --> 00:51:14,180
I just have reorient to put
middle bars in the xy plane.

1023
00:51:14,180 --> 00:51:17,280
So in other words,
there's two middle bars.

1024
00:51:17,280 --> 00:51:20,140
They live in a plane
because they share a vertex.

1025
00:51:20,140 --> 00:51:22,570
So think of that as the plane
and then scale perpendicular

1026
00:51:22,570 --> 00:51:23,530
to the that.

1027
00:51:23,530 --> 00:51:26,150
Same trick works.

1028
00:51:26,150 --> 00:51:29,162
So 4 chains versus
various 2 chains.

1029
00:51:29,162 --> 00:51:31,370
Again, we preserve the
lengths of the two middle bars

1030
00:51:31,370 --> 00:51:32,740
because they lie on the plane.

1031
00:51:32,740 --> 00:51:34,480
So there's no scaling
that happens in the plane,

1032
00:51:34,480 --> 00:51:35,910
and in the generic
case, everybody

1033
00:51:35,910 --> 00:51:37,409
will be off the
plane, and they fly.

1034
00:51:39,680 --> 00:51:41,009
Cool.

1035
00:51:41,009 --> 00:51:43,050
So that proves all of
these separability results.

1036
00:51:43,050 --> 00:51:46,167
I'm not going to do these
closed chain ones here.

1037
00:51:46,167 --> 00:51:48,000
You can read the paper
if you're interested.

1038
00:51:48,000 --> 00:51:50,280
At this point, we can go
back to Lubiw's problem

1039
00:51:50,280 --> 00:51:54,520
and prove this result.

1040
00:51:54,520 --> 00:51:58,569
So basic algorithm is
cut every other vertex.

1041
00:51:58,569 --> 00:52:00,110
You'll get a whole
bunch of hairpins.

1042
00:52:00,110 --> 00:52:02,130
Hairpins, we know, separate.

1043
00:52:02,130 --> 00:52:04,380
But we can be slightly
more efficient

1044
00:52:04,380 --> 00:52:09,405
by leaving an initial segment
of 4 chain or two 3 chains.

1045
00:52:09,405 --> 00:52:12,900
I think it comes out
to the same number.

1046
00:52:12,900 --> 00:52:18,855
And you end up saving three
if we computed this right.

1047
00:52:18,855 --> 00:52:21,830
So you get n minus 3 over 2.

1048
00:52:21,830 --> 00:52:24,490
I think our first bound on this
problem was n minus 1 over 2,

1049
00:52:24,490 --> 00:52:27,250
and then we improved by 1
although I have to n minus 3

1050
00:52:27,250 --> 00:52:28,200
over 2.

1051
00:52:28,200 --> 00:52:32,090
That has stood the
best bound since.

1052
00:52:32,090 --> 00:52:35,110
Basically, because with
interlocked stuff-- after that,

1053
00:52:35,110 --> 00:52:36,310
you get interlocked.

1054
00:52:36,310 --> 00:52:38,534
And if you're going
to avoid interlocking,

1055
00:52:38,534 --> 00:52:39,450
you have to be clever.

1056
00:52:39,450 --> 00:52:42,500
You can't just use the numbers.

1057
00:52:42,500 --> 00:52:44,720
So that is all we know
about Lubiw's problem.

1058
00:52:44,720 --> 00:52:49,470
Now, I get to show you all
the fun interlocked examples.

1059
00:52:49,470 --> 00:52:54,380
We're going to do pretty
much all of these.

1060
00:52:54,380 --> 00:52:59,580
So we start with one of
the prettiest examples,

1061
00:52:59,580 --> 00:53:00,610
the threefold symmetric.

1062
00:53:00,610 --> 00:53:04,380
This is three 3 chains.

1063
00:53:04,380 --> 00:53:08,190
So that's not even drawn here.

1064
00:53:08,190 --> 00:53:11,380
Because this table is about
pairwise interactions.

1065
00:53:11,380 --> 00:53:14,480
We know that two 3 chains-- we
just proved two 3 chains cannot

1066
00:53:14,480 --> 00:53:18,400
interlock, but
three of them can.

1067
00:53:18,400 --> 00:53:20,765
That's the only case where
that question is interesting.

1068
00:53:20,765 --> 00:53:23,490
So everybody else,
you can interlock.

1069
00:53:23,490 --> 00:53:26,120
Unless you have 2 chains,
and we know those don't help.

1070
00:53:26,120 --> 00:53:28,534
So three 3 chains interlock.

1071
00:53:28,534 --> 00:53:30,200
I'm not going to try
to prove that here.

1072
00:53:30,200 --> 00:53:32,170
It's a pretty
complicated argument

1073
00:53:32,170 --> 00:53:35,149
where you look at
the center edges.

1074
00:53:35,149 --> 00:53:36,940
You look at the convex
whole of those edges

1075
00:53:36,940 --> 00:53:38,981
and argue these guys can't
stick into the center.

1076
00:53:38,981 --> 00:53:40,715
It's a geometric argument.

1077
00:53:40,715 --> 00:53:43,385
If you're interested,
check out this paper.

1078
00:53:43,385 --> 00:53:44,820
It's a pretty cool example.

1079
00:53:44,820 --> 00:53:47,160
This was a great-- finally,
I had a good exercise

1080
00:53:47,160 --> 00:53:49,717
for using Rhino yesterday
and drawing this figure.

1081
00:53:49,717 --> 00:53:50,550
It was a lot of fun.

1082
00:53:54,370 --> 00:53:57,510
This is a 3 chain and 4 chain.

1083
00:53:57,510 --> 00:53:59,840
One, two, three, four.

1084
00:53:59,840 --> 00:54:01,920
Just two different
views of the same thing.

1085
00:54:01,920 --> 00:54:07,000
And that proves this result,
which is this result.

1086
00:54:11,780 --> 00:54:15,350
It's pretty close, pretty tight.

1087
00:54:15,350 --> 00:54:17,140
Again, it's a
geometric argument.

1088
00:54:17,140 --> 00:54:18,050
It's tricky.

1089
00:54:18,050 --> 00:54:19,110
I'll go into it here.

1090
00:54:22,770 --> 00:54:24,830
We found these by building
lots of straw models

1091
00:54:24,830 --> 00:54:26,840
and jiggling them
around with string

1092
00:54:26,840 --> 00:54:28,860
connecting the straws together.

1093
00:54:28,860 --> 00:54:30,436
But then we had to prove it.

1094
00:54:30,436 --> 00:54:32,435
For a while, we weren't
sure whether a 3 and a 4

1095
00:54:32,435 --> 00:54:33,340
could interlock.

1096
00:54:33,340 --> 00:54:35,910
I think first we had a 4
and a 4 or maybe 4 and a 5,

1097
00:54:35,910 --> 00:54:37,490
and we kept working
our way down.

1098
00:54:37,490 --> 00:54:39,531
And now we know this is
the smallest you can get.

1099
00:54:42,460 --> 00:54:47,650
And now we go into
the closed results.

1100
00:54:47,650 --> 00:54:50,690
Because 4 and 4, that's of
course easier than 3 and 4.

1101
00:54:50,690 --> 00:54:53,040
So that's done.

1102
00:54:53,040 --> 00:54:56,940
This is contained
in those results.

1103
00:54:56,940 --> 00:54:59,020
Now, we look at the
closed examples.

1104
00:54:59,020 --> 00:55:02,980
So this is going to be a
closed triangle interlocking

1105
00:55:02,980 --> 00:55:04,627
with an open 4 chain.

1106
00:55:04,627 --> 00:55:06,710
And here I'm going to give
you some of the details

1107
00:55:06,710 --> 00:55:09,822
because this proof-- it's
relatively easy to prove

1108
00:55:09,822 --> 00:55:10,780
that this is interlock.

1109
00:55:10,780 --> 00:55:12,490
You can use a more
topological argument,

1110
00:55:12,490 --> 00:55:14,740
which is what we're used to
from the knitting needles.

1111
00:55:14,740 --> 00:55:18,200
Remember we took a ball that
separated the interior vertices

1112
00:55:18,200 --> 00:55:20,230
from the endpoints.

1113
00:55:20,230 --> 00:55:23,080
We said the endpoints
really can't get in

1114
00:55:23,080 --> 00:55:24,420
if we set the ball right.

1115
00:55:24,420 --> 00:55:29,550
And therefore, you could connect
the endpoints by a big rope,

1116
00:55:29,550 --> 00:55:31,816
and then you have
an Trefoil knot,

1117
00:55:31,816 --> 00:55:33,940
and there's no way you can
untie the knot with that

1118
00:55:33,940 --> 00:55:35,240
having intersection somewhere.

1119
00:55:35,240 --> 00:55:37,150
You can argue the rope
never self intersects,

1120
00:55:37,150 --> 00:55:40,390
so it must be the chain.

1121
00:55:40,390 --> 00:55:42,510
We can do the same
thing with this example,

1122
00:55:42,510 --> 00:55:46,870
with a 4 chain interlocked
with a triangle.

1123
00:55:46,870 --> 00:55:53,160
And we'll sketch that proof.

1124
00:56:17,300 --> 00:56:19,830
So I want to prove that those
two guys are interlocked,

1125
00:56:19,830 --> 00:56:23,210
so suppose that somehow they
separate from each other.

1126
00:56:23,210 --> 00:56:26,175
If there's a separating
motion, there's

1127
00:56:26,175 --> 00:56:28,970
a motion where the triangle
actually doesn't move at all.

1128
00:56:28,970 --> 00:56:30,980
So triangle's rigid,
except that it

1129
00:56:30,980 --> 00:56:34,670
can move by rigid motion,
translation and rotation.

1130
00:56:34,670 --> 00:56:39,420
But any motion I have, I can
recenter things by relativity

1131
00:56:39,420 --> 00:56:41,570
so that the
triangles stays fixed

1132
00:56:41,570 --> 00:56:45,114
and just the
quadrilateral moves.

1133
00:56:45,114 --> 00:56:47,280
So that's nice because a
triangle is a natural thing

1134
00:56:47,280 --> 00:56:49,230
to draw a sphere
around, and I want

1135
00:56:49,230 --> 00:56:52,910
to argue if these bars
are really, really long,

1136
00:56:52,910 --> 00:56:56,920
the endpoints will be
outside this big sphere.

1137
00:56:56,920 --> 00:56:58,200
So let me tell you how long.

1138
00:57:02,320 --> 00:57:04,775
I have my triangle
I'm going to draw

1139
00:57:04,775 --> 00:57:07,350
a sphere that
contains the triangle.

1140
00:57:07,350 --> 00:57:09,830
Say this sphere has
radius little r.

1141
00:57:09,830 --> 00:57:12,680
Big R is going to be
little r plus the sum

1142
00:57:12,680 --> 00:57:16,210
of the middle bars, which
should sound familiar from what

1143
00:57:16,210 --> 00:57:18,180
we did with knitting needles.

1144
00:57:18,180 --> 00:57:20,614
This is of the 4 chain.

1145
00:57:20,614 --> 00:57:22,340
There's two middle bars.

1146
00:57:22,340 --> 00:57:24,090
Add up their lengths.

1147
00:57:24,090 --> 00:57:24,830
Add on r.

1148
00:57:24,830 --> 00:57:27,620
This is just a big
number, capital R.

1149
00:57:27,620 --> 00:57:29,770
I'm going to make it bigger.

1150
00:57:29,770 --> 00:57:32,030
So I have my little triangle.

1151
00:57:32,030 --> 00:57:34,290
It has a tiny ball around it.

1152
00:57:34,290 --> 00:57:38,130
I'm going to draw a
pretty big sphere here

1153
00:57:38,130 --> 00:57:46,789
of radius 15 r, an even
bigger sphere of radius 20 r.

1154
00:57:46,789 --> 00:57:49,330
And I don't need to be precise
about exactly where the center

1155
00:57:49,330 --> 00:57:51,496
is, but I guess it would
be the center of this ball.

1156
00:57:55,630 --> 00:57:57,930
Here's what I want to say.

1157
00:57:57,930 --> 00:58:04,320
I'm supposing the lengths of
these edges is more than 20 r.

1158
00:58:04,320 --> 00:58:08,430
So initially, those
bars go outside the box.

1159
00:58:08,430 --> 00:58:10,500
The endpoints are
outside of the big ball.

1160
00:58:13,060 --> 00:58:19,030
Now, here's the claim.

1161
00:58:19,030 --> 00:58:22,570
As long as the endpoints of
the really long-- the endpoints

1162
00:58:22,570 --> 00:58:35,280
of the 4 chain stay outside this
ball, then the interior bars--

1163
00:58:35,280 --> 00:58:42,760
these two middle bars-- have
to stay inside this ball.

1164
00:58:42,760 --> 00:58:44,760
This is just like
the knitting needles.

1165
00:58:44,760 --> 00:58:47,170
As long as the super
long bars, the endpoints,

1166
00:58:47,170 --> 00:58:49,730
stay outside this
ball, the middle bars

1167
00:58:49,730 --> 00:58:50,900
have to stay inside.

1168
00:58:50,900 --> 00:58:54,480
That's just because those
bars are so darn long.

1169
00:58:54,480 --> 00:58:57,000
And so for as long
as that holds,

1170
00:58:57,000 --> 00:58:59,720
you can compute the
topology-- I'm not

1171
00:58:59,720 --> 00:59:05,920
going to try to draw it-- the
topology of-- in this picture,

1172
00:59:05,920 --> 00:59:08,200
if you connect to the
ends of the purple guys,

1173
00:59:08,200 --> 00:59:12,419
because they're outside the
big ball, the 15 r ball--

1174
00:59:12,419 --> 00:59:13,960
not sure I really
need the 20 r ball,

1175
00:59:13,960 --> 00:59:19,310
but the 15 r ball is the key--
topology you get is this.

1176
00:59:19,310 --> 00:59:22,190
That is the link
between the circle,

1177
00:59:22,190 --> 00:59:23,800
which is this nice loop.

1178
00:59:23,800 --> 00:59:27,220
And the purple guy
is that red loop

1179
00:59:27,220 --> 00:59:29,490
if you just figure out what
it looks like currently.

1180
00:59:29,490 --> 00:59:33,710
And as long as the endpoints
of these guys don't go inside

1181
00:59:33,710 --> 00:59:35,420
and those guys
remain on the inside,

1182
00:59:35,420 --> 00:59:37,570
basically until something
interesting happens,

1183
00:59:37,570 --> 00:59:38,840
the topology has to be that.

1184
00:59:41,810 --> 00:59:49,110
So at some point-- sorry,
I said that slightly wrong.

1185
00:59:49,110 --> 00:59:52,250
As long as the middle
bars stay inside the ball

1186
00:59:52,250 --> 00:59:55,920
and the endpoint stay
outside the ball, a 15 r,

1187
00:59:55,920 --> 00:59:56,949
this will be the case.

1188
00:59:56,949 --> 00:59:58,990
They are not saying nothing
interesting happened.

1189
00:59:58,990 --> 01:00:01,490
So violation, the
only way to open

1190
01:00:01,490 --> 01:00:06,340
this thing could be an endpoint
goes inside or a middle bar

1191
01:00:06,340 --> 01:00:09,340
goes outside the ball.

1192
01:00:09,340 --> 01:00:12,630
Consider the moment
when that happens.

1193
01:00:12,630 --> 01:00:16,810
If a middle bar, one
in the middle vertices

1194
01:00:16,810 --> 01:00:21,330
is going to go outside
this 15 r radius ball,

1195
01:00:21,330 --> 01:00:25,520
then you know that
all of the middle bars

1196
01:00:25,520 --> 01:00:29,390
are far away from this
triangle because those edges

1197
01:00:29,390 --> 01:00:32,670
very small compared
to capital R.

1198
01:00:32,670 --> 01:00:34,860
So we're 15 away.

1199
01:00:34,860 --> 01:00:37,760
If you have one middle vertex
that's outside, all of them

1200
01:00:37,760 --> 01:00:39,000
are pretty close to outside.

1201
01:00:39,000 --> 01:00:41,070
They're all going to
be in some smaller ball

1202
01:00:41,070 --> 01:00:45,720
here-- outside
that smaller ball.

1203
01:00:45,720 --> 01:00:48,530
The alternative is that
one of the end-- that's

1204
01:00:48,530 --> 01:00:50,727
not a contradiction yet,
but we'll come back to it--

1205
01:00:50,727 --> 01:00:53,060
the alternative is that one
of the endpoints of the bars

1206
01:00:53,060 --> 01:00:55,440
goes inside this
radius 15 r ball.

1207
01:00:58,280 --> 01:01:02,180
When that happens, because
this bar is so long,

1208
01:01:02,180 --> 01:01:06,250
the other end of the bar will
actually be outside the sphere.

1209
01:01:06,250 --> 01:01:08,575
And so then, again,
all the middle guys

1210
01:01:08,575 --> 01:01:10,390
are outside the sphere.

1211
01:01:10,390 --> 01:01:12,100
So either way, you violate.

1212
01:01:12,100 --> 01:01:13,730
I assumed this
before, but now, I'm

1213
01:01:13,730 --> 01:01:16,020
saying if you try to
take these guys out

1214
01:01:16,020 --> 01:01:17,820
or if you try to
push this guy in,

1215
01:01:17,820 --> 01:01:20,480
the middle guys are
outside the ball already.

1216
01:01:20,480 --> 01:01:23,870
So consider a situation when
the middle edges are all outside

1217
01:01:23,870 --> 01:01:26,490
the ball of radius 15 r.

1218
01:01:26,490 --> 01:01:29,720
The only thing that could
penetrate some tiny sphere,

1219
01:01:29,720 --> 01:01:33,530
like this one of little r,
only parts of the 4 chain

1220
01:01:33,530 --> 01:01:35,800
that could penetrate
that are the long bars

1221
01:01:35,800 --> 01:01:38,450
because the middle
stuff is too far away.

1222
01:01:38,450 --> 01:01:41,800
The long bars are super long,
so they could conceivably

1223
01:01:41,800 --> 01:01:43,430
penetrate the triangle.

1224
01:01:43,430 --> 01:01:45,510
All we really care about
is piercing the center

1225
01:01:45,510 --> 01:01:46,968
of the triangle
because that's when

1226
01:01:46,968 --> 01:01:48,460
we get interesting topology.

1227
01:01:48,460 --> 01:01:52,370
We want to get this
topology, but sadly, you

1228
01:01:52,370 --> 01:01:54,050
could only get these topologies.

1229
01:01:54,050 --> 01:01:54,804
I guess luckily.

1230
01:01:54,804 --> 01:01:56,470
You can only get those
three topologies,

1231
01:01:56,470 --> 01:01:58,060
which don't match this one.

1232
01:01:58,060 --> 01:01:59,780
And so there's no way
to get continuously

1233
01:01:59,780 --> 01:02:01,890
from one of these to the other.

1234
01:02:01,890 --> 01:02:02,900
That's the argument.

1235
01:02:02,900 --> 01:02:04,930
Let me say a little
more which is

1236
01:02:04,930 --> 01:02:08,360
if neither of the two long
bars hit the triangle,

1237
01:02:08,360 --> 01:02:10,770
then you trivially get this.

1238
01:02:10,770 --> 01:02:12,860
You don't get any linking.

1239
01:02:12,860 --> 01:02:16,830
If one of them strikes a
triangle, you always get this.

1240
01:02:19,430 --> 01:02:21,640
If two of them
strike the triangle,

1241
01:02:21,640 --> 01:02:23,510
it's possible to
get this or this

1242
01:02:23,510 --> 01:02:24,902
depending on how you link it up.

1243
01:02:24,902 --> 01:02:26,360
And I'm not going
to prove it here,

1244
01:02:26,360 --> 01:02:27,910
but that's all you can get.

1245
01:02:27,910 --> 01:02:31,875
And so if you can't get
this one, you're in trouble.

1246
01:02:31,875 --> 01:02:34,850
There's no continuous
motion from this to those

1247
01:02:34,850 --> 01:02:36,604
without crossing.

1248
01:02:36,604 --> 01:02:37,520
So that's how it goes.

1249
01:02:37,520 --> 01:02:40,850
And this is an example of
a topological argument.

1250
01:02:40,850 --> 01:02:42,710
You can use the same
kind of argument

1251
01:02:42,710 --> 01:02:48,310
to prove this last
result, which is this one.

1252
01:02:48,310 --> 01:02:51,220
But now this is not necessarily
symmetric because one's closed,

1253
01:02:51,220 --> 01:02:52,370
one's open.

1254
01:02:52,370 --> 01:02:57,340
So this is a closed 4 chain
quadrilateral, in other words,

1255
01:02:57,340 --> 01:03:01,150
with an open 3 chain,
which can also lock.

1256
01:03:01,150 --> 01:03:03,340
And again, you can use
the topological arguments.

1257
01:03:03,340 --> 01:03:05,881
A little messy because now this
is like a tetrahedron instead

1258
01:03:05,881 --> 01:03:10,510
of a triangle, but the same
tricks work and topology,

1259
01:03:10,510 --> 01:03:13,590
luckily, doesn't match between
the inside and outside cases.

1260
01:03:16,350 --> 01:03:17,650
Cool.

1261
01:03:17,650 --> 01:03:22,740
So that is interlocking
and pretty much

1262
01:03:22,740 --> 01:03:24,390
the entirety of this table.

1263
01:03:24,390 --> 01:03:27,070
Now, there's a way to
generalize this table,

1264
01:03:27,070 --> 01:03:29,800
which is instead of saying,
oh, I have universal joints.

1265
01:03:29,800 --> 01:03:31,510
Everything's nice.

1266
01:03:31,510 --> 01:03:32,930
I start thinking
back to proteins

1267
01:03:32,930 --> 01:03:35,441
and say, well, what if
I had some fixed angles.

1268
01:03:35,441 --> 01:03:37,690
What if some of the chains
are fixed angles and others

1269
01:03:37,690 --> 01:03:38,540
are universal?

1270
01:03:38,540 --> 01:03:40,170
What if some of the
chains were ridge?

1271
01:03:40,170 --> 01:03:42,890
Because we know that rigid
hairpins can separate

1272
01:03:42,890 --> 01:03:45,610
without any folding whatsoever,
what if some of them are rigid

1273
01:03:45,610 --> 01:03:48,590
and some of them are fixed angle
and some of them are universal?

1274
01:03:48,590 --> 01:03:53,320
Well, we haven't done all of
those results, but most of them

1275
01:03:53,320 --> 01:03:58,070
for pairs, we have got
universal and rigid.

1276
01:03:58,070 --> 01:04:01,130
I mean a fixed angle 2
chain is a rigid 2 chain,

1277
01:04:01,130 --> 01:04:02,970
so there's only
two columns there.

1278
01:04:02,970 --> 01:04:04,160
Universal fixed angle rigid.

1279
01:04:04,160 --> 01:04:05,720
Universal fixed angle rigid.

1280
01:04:05,720 --> 01:04:07,650
And here we only
get up to rigid.

1281
01:04:11,000 --> 01:04:14,400
Yeah, I'm not sure if it makes
a huge difference out there.

1282
01:04:14,400 --> 01:04:17,120
Again, this is not
a complete story.

1283
01:04:17,120 --> 01:04:20,860
There other results
that might fit in here.

1284
01:04:20,860 --> 01:04:25,120
How far down do those minuses go
if you don't have rigid parts?

1285
01:04:25,120 --> 01:04:27,300
But let's see.

1286
01:04:27,300 --> 01:04:28,890
The plus means
there's locked things.

1287
01:04:28,890 --> 01:04:32,070
The minus means they
always separate.

1288
01:04:32,070 --> 01:04:33,720
It's a little more concise.

1289
01:04:33,720 --> 01:04:35,870
And we've sorted
the rows and columns

1290
01:04:35,870 --> 01:04:38,820
so that you get nice--
conveniently you

1291
01:04:38,820 --> 01:04:42,620
get a nice diagonal pattern,
not perfect diagonal,

1292
01:04:42,620 --> 01:04:45,430
but it's at least monotone.

1293
01:04:45,430 --> 01:04:48,240
It makes sense that a
rigid-- obviously, rigid

1294
01:04:48,240 --> 01:04:52,690
is more fixed than
a fixed angle.

1295
01:04:52,690 --> 01:04:54,940
But it's not obvious
that a universal 4 chain

1296
01:04:54,940 --> 01:04:57,380
is going to lock more than
a fixed angle 3 chain,

1297
01:04:57,380 --> 01:05:01,630
but turns out that is the
case for these examples.

1298
01:05:01,630 --> 01:05:03,749
Now, you could consider
triples of them,

1299
01:05:03,749 --> 01:05:05,040
each with different properties.

1300
01:05:05,040 --> 01:05:06,300
Then, you get a three
dimensional table.

1301
01:05:06,300 --> 01:05:07,560
That gets a little messier.

1302
01:05:07,560 --> 01:05:08,930
Probably not a huge
number of results

1303
01:05:08,930 --> 01:05:10,290
are necessary along those lines.

1304
01:05:10,290 --> 01:05:16,630
Here, we just needed the 2
chains plus the three 3 chains

1305
01:05:16,630 --> 01:05:18,350
were interlocked.

1306
01:05:18,350 --> 01:05:20,330
So there's more work
to be done here,

1307
01:05:20,330 --> 01:05:22,790
but this was a lot of fun.

1308
01:05:22,790 --> 01:05:24,290
And if you want to
see the examples,

1309
01:05:24,290 --> 01:05:26,020
you should check out this paper.

1310
01:05:30,530 --> 01:05:32,960
This was basically
for fun and to see how

1311
01:05:32,960 --> 01:05:34,940
low could you go in
terms of number of edges

1312
01:05:34,940 --> 01:05:37,190
and various combinations
of different parts.

1313
01:05:37,190 --> 01:05:40,831
But it turns out this interlock
stuff does have applications.

1314
01:05:40,831 --> 01:05:42,330
I'm not going to
call them practical

1315
01:05:42,330 --> 01:05:44,680
because it's about
another theory result.

1316
01:05:44,680 --> 01:05:47,060
And its result, I mentioned.

1317
01:05:47,060 --> 01:05:50,100
If you have some 3
chain-- oh sorry.

1318
01:05:50,100 --> 01:05:53,990
Some three dimensional chain--
let's say universal joints--

1319
01:05:53,990 --> 01:05:57,380
and you have two
configurations, configuration a

1320
01:05:57,380 --> 01:06:01,270
and configuration
b, and you want

1321
01:06:01,270 --> 01:06:03,415
to know is there a
motion from configuration

1322
01:06:03,415 --> 01:06:06,620
a to configuration b.

1323
01:06:06,620 --> 01:06:16,800
This is PSPACE-complete,
result by a bunch of Berliners

1324
01:06:16,800 --> 01:06:18,120
from a few years ago.

1325
01:06:18,120 --> 01:06:21,000
And basically, what they
do is build a computer.

1326
01:06:21,000 --> 01:06:24,680
I don't have the gadgets
here to show you.

1327
01:06:24,680 --> 01:06:27,500
But they build a computer
with various moving parts

1328
01:06:27,500 --> 01:06:28,780
and what not.

1329
01:06:28,780 --> 01:06:31,230
To build the infrastructure
of that computer,

1330
01:06:31,230 --> 01:06:33,490
the rigid part doesn't move.

1331
01:06:33,490 --> 01:06:35,610
They use interlocked chains.

1332
01:06:35,610 --> 01:06:39,540
Basically, if you want to
build a nice, rigid framework,

1333
01:06:39,540 --> 01:06:44,497
like a truss, you can do it
because you just take chains.

1334
01:06:44,497 --> 01:06:45,830
I'm only allowed to make chains.

1335
01:06:45,830 --> 01:06:47,121
I'm not allowed to make graphs.

1336
01:06:47,121 --> 01:06:50,110
But if I bring two
parts of the chain

1337
01:06:50,110 --> 01:06:52,320
really close to each
other, I can basically

1338
01:06:52,320 --> 01:06:55,300
tie a knot by making
lots of little links

1339
01:06:55,300 --> 01:06:57,980
and using say an
interlocked 3, 4 chain,

1340
01:06:57,980 --> 01:06:59,922
interlock 3 chain with
an interlocked 4 chain.

1341
01:06:59,922 --> 01:07:01,380
In the same way
when we're building

1342
01:07:01,380 --> 01:07:06,640
the locked 11 chain
with the lock 2 chain,

1343
01:07:06,640 --> 01:07:09,362
you're tying little knots
around various points.

1344
01:07:09,362 --> 01:07:11,070
Here, we don't even
have to be efficient,

1345
01:07:11,070 --> 01:07:13,190
just constant numbers are fine.

1346
01:07:13,190 --> 01:07:19,020
So you can build some rigid
structure like a tetrahedron

1347
01:07:19,020 --> 01:07:27,200
by taking one path here,
making an interlock structure,

1348
01:07:27,200 --> 01:07:31,290
making interlock structure,
interlock, interlock.

1349
01:07:31,290 --> 01:07:34,930
Could double traverse, I guess.

1350
01:07:34,930 --> 01:07:38,100
And you can interlock
everything using a chain.

1351
01:07:38,100 --> 01:07:39,632
Build a rigid structure.

1352
01:07:39,632 --> 01:07:41,340
Then, you go and build
the flexible part.

1353
01:07:41,340 --> 01:07:43,250
And that's the fun computation.

1354
01:07:43,250 --> 01:07:45,590
But just to get off
the ground and to build

1355
01:07:45,590 --> 01:07:49,160
some rigid infrastructure,
to build gadgets basically

1356
01:07:49,160 --> 01:07:51,575
in your hardness proof,
they use interlocked chains,

1357
01:07:51,575 --> 01:07:52,450
which is pretty cool.

1358
01:07:52,450 --> 01:07:57,310
This is a really interesting
result, a very negative one.

1359
01:07:57,310 --> 01:07:59,480
It's really hard to fold
from one shape to another,

1360
01:07:59,480 --> 01:08:01,290
but it uses this.

1361
01:08:01,290 --> 01:08:03,770
They also prove the same thing
for trees in two dimensions.

1362
01:08:03,770 --> 01:08:05,270
For that, they used
the lock trees--

1363
01:08:05,270 --> 01:08:07,270
which we covered way
back, lecture three

1364
01:08:07,270 --> 01:08:11,720
or whatever-- these lock trees
to build-- to basically fill

1365
01:08:11,720 --> 01:08:15,670
rigid angles, to force things
to be at particular angles,

1366
01:08:15,670 --> 01:08:19,520
and then they have flexible
parts to do the computer stuff.

1367
01:08:19,520 --> 01:08:20,140
So it's fun.

1368
01:08:20,140 --> 01:08:24,370
They use our little examples
to build big computers.

1369
01:08:24,370 --> 01:08:27,660
Still open, of
course, is can I start

1370
01:08:27,660 --> 01:08:32,120
from a straight
configuration of the chain

1371
01:08:32,120 --> 01:08:34,790
and can I get it to this.

1372
01:08:34,790 --> 01:08:38,500
So in other words, given
a chain, can I unfold it.

1373
01:08:38,500 --> 01:08:40,910
We'd like to know whether
that's polynomially

1374
01:08:40,910 --> 01:08:42,200
solvable or empty complete.

1375
01:08:42,200 --> 01:08:45,670
It is about characterizing
lock chains in some sense.

1376
01:08:45,670 --> 01:08:47,122
This is about going
from arbitrary

1377
01:08:47,122 --> 01:08:48,830
configuration to
arbitrary configuration.

1378
01:08:48,830 --> 01:08:50,649
And here, you're
allowed to build

1379
01:08:50,649 --> 01:08:53,109
rigid infrastructure
which never moves.

1380
01:08:53,109 --> 01:08:55,620
Here, everything would
have to fall apart somehow,

1381
01:08:55,620 --> 01:08:57,779
so you can't use
interlock chains.

1382
01:08:57,779 --> 01:09:00,790
You can't use lock trees in 2D.

1383
01:09:00,790 --> 01:09:06,285
So this problem is still open.

1384
01:09:06,285 --> 01:09:07,660
Seems quite
difficult because you

1385
01:09:07,660 --> 01:09:11,149
can't build any infrastructure
the stays around forever.

1386
01:09:11,149 --> 01:09:13,020
Everything would
have to fall apart

1387
01:09:13,020 --> 01:09:14,195
if your machine terminates.

1388
01:09:16,779 --> 01:09:19,181
Any questions?

1389
01:09:19,181 --> 01:09:20,930
Wanted to talk about
that for a long time.

1390
01:09:20,930 --> 01:09:22,346
Now that we have
interlock chains,

1391
01:09:22,346 --> 01:09:24,979
you get to see a little bit
behind the scenes of what's

1392
01:09:24,979 --> 01:09:25,689
happening.

1393
01:09:25,689 --> 01:09:27,200
If you want to see
the full proof,

1394
01:09:27,200 --> 01:09:32,370
check out that paper by
[INAUDIBLE], [INAUDIBLE],

1395
01:09:32,370 --> 01:09:33,859
and [INAUDIBLE].

1396
01:09:38,290 --> 01:09:38,810
All right.

1397
01:09:38,810 --> 01:09:40,570
That's it.