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PROFESSOR: Today
we're going to talk

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about protein folding and its
relation to linkage folding.

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We're going to look at a
mechanical model of proteins.

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This is an example of a
protein from lecture one.

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There's a ton out
there in this place

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called the protein data
bank, all freely available.

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It's really hard to
get pictures like this.

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But you get some
idea that there's

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a linkage embedded in here.

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You see various little
spheres and edges.

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That's, of course, not reality.

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Those spheres are
actually atoms and they're

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kind of amorphous blobs.

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The edges are chemical bonds.

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And those are connections.

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We don't know whether they're--
it's not really matter,

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but it's force.

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This is a rather messy picture.

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This is what a protein
folds into, some 3D shape.

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Most proteins fold
consistently into one shape.

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We don't really know
how that happens.

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We can't watch it happen.

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So the big challenge is
to know how proteins fold.

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Given a protein, what
does it fold into?

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That's the protein
folding problem.

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Major unsolved problem
in biology, biochemistry.

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The protein design problem is
I want to make a particular 3D

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shape so that it docks into
something, binds to a virus,

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whatever, what protein should
I synthesize in order for it

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to fold into that shape?

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That is potentially an easier
question algorithmically

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and it's the really useful one
from a drug design standpoint.

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Some new virus comes
along, you design a drug

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to attack it and only
it, you build it.

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Usually you would manufacture
some synthetic DNA,

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you feed it into the
cell, DNA goes to the RNA,

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goes to the mRNA,
goes to the protein.

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You all remember
biology 101, hopefully.

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We don't need to
know much about it.

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If you look at what's called
the backbone of the protein,

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a protein's basically a chain
and attached to the chain

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are various amino acids.

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Today I'm going to ignore
the amino acids, which

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is a little crazy, and just
think about the backbone chain.

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Backbone chain looks
something like this.

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One of the challenges of
video recording a class is I

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can only use copyright free
or Creative Commons images.

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This one, I couldn't get
one, so I had to draw it.

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There's various
measurements here

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in certain numbers of angstroms.

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Those are the chemical bonds.

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Various atoms here-- nitrogen,
carbon, and so on, hydrogen.

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But basically, it's a chain
that zigzags back and forth.

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You can also see
the angles here.

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They're not quite all the
same, but they're very similar.

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All the lengths in
the angles are close.

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It zigzags-- this is
really in three dimensions.

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I tried to draw the spheres
so you could see the three

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dimensionality, but
it's a little tricky.

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And then attached on the
sides are the amino acids.

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I'm going to focus
just on the backbone.

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The way this thing is allowed
to fold-- these lengths,

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as far as we know,
are pretty static.

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They probably would
jiggle a little bit.

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But you can think
of them as edges.

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So you can think of
this as a linkage.

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The catch is, also,
the angles are

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fixed because the way this atom
wants to bind to other things

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has very fixed angle patterns.

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If you ever played with a
chemistry construction set,

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that's how they work.

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They have holes at
just particular angles.

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So if you think of
like a robotic arm,

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normally-- like here, I have
it two edge robotic arm,

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let's say.

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Normally, you have two degrees
of freedom in three dimensions.

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You can change the angle and
you can spin around this edge.

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Now, it's saying the angle
is fixed-- for example,

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here it's, say, at 90 degrees.

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All I can do is spin.

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I'm not allowed to flex
my muscle in this way.

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So that is the model.

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All of-- in this case, we have
a tree-- all of the angles

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here are fixed.

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But you can still, for example,
take this entire sub chain

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and spin it around this edge.

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That'll preserve all the
angles and all the lengths.

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That's all you're allowed to do.

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You take an edge,
you spin it-- spin

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one half of the edge
relative to the other half.

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These are called
fixed angle linkages.

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And they have been studied
quite a lot because

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of their connection
to protein folding.

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So embedded in the
term linkage, we

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assume that the edge
lengths are fixed,

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and then we add the constraint
that the angles are fixed.

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And the motivation
is the backbone

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is something like-- the
backbone of a protein

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is something like
a fixed angle tree.

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Of course, it's
not much of a tree.

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Most of it is a chain.

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There's just small
objects hanging off,

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and if you add the
amino acid there

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are bigger things hanging
off, but still constant size.

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They'll have some cycles.

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They're not trees.

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But it's slightly
more approximately--

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I should draw wavier lines.

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It's a chain.

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Usually an open chain although
occasionally a closed chain.

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So we think a lot about fixed
angle chain and sometimes about

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fixed angle trees.

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Now, fixed angle linkages
are harder to think

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about than universal
joints-- that's

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the usual kind of linkage.

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So we know 3D linkages
are kind of tough.

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Nonetheless, we found
lots of really interesting

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mathematical problems
to solve here,

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and that is the topic of today.

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At some level, we are
thinking about the mechanics

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of protein folding.

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We're throwing away energy.

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We're throwing
away the actuators

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in real life that
make proteins fold.

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We're just imagining,
given this mechanical model

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of how a protein might
fold, what's possible.

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So in some sense, it's
broader than reality.

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And the hope is you find
an interesting algorithm

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for how to fold
these protein chains.

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And maybe that's the algorithm
that nature is implementing.

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That's the kind of
general picture.

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We're not constrained
by reality,

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and by how nature
actually folds things.

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So I'm going to talk today
about four main problems here.

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The first one's called span.

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Second one's called flattening.

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Third one is flat
state connectivity.

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And the fourth one is locked,
our good friend locked chains.

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And of course there
are locked chains

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because we're constraining
linkages even more than before.

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So you could take knitting
needles, it'll still be locked.

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So you add extra constraints.

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Makes it harder to fold.

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But there are actually some
interesting positive results

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we can give of chains that
are not locked in some sense.

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And flat state connectivity is
about the same kind of thing,

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where instead of worrying
about getting from anywhere

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to anywhere, we just
worry about getting

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from one flat state
to another flat state.

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Flat means lying in a plane.

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Flattening is about is
there such a configuration.

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And span is about
given robotic arm--

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like a more complicated one,
like with multiple edges-- how

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far apart can the endpoints
get, and how close

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can the endpoints get.

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The universal chain
is not very exciting.

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Farthest it can get is when
it's straight, and the least far

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it can get is when it's closed.

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You can always do that, I think.

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Well, no, I guess you
can't always close it up.

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That's a little nontrivial.

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But for fixed
angle linkages, you

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can't straighten out because
you have to preserve the angles.

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So it's kind of what is the
straightest like configuration,

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given that the angles are fixed.

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So let's start with span.

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So the span of a
configuration is the distance

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between the endpoints.

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And in general, you'll find
the max span and the min span.

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This search was
begun by a guy named

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Mike Soss, who was a
PhD student at McGill.

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And he proved that
if you want to find,

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for example, a flat
state that lives

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in two dimensions with the
minimum or the maximum span,

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this is NP-hard.

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This is in his PhD thesis.

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Question?

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AUDIENCE: If you have a
linkage or [INAUDIBLE]

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chain that actually loops
around, is there a span

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because there is no endpoint?

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PROFESSOR: Oh, here I'm
assuming open chain--

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I should say that--
which most proteins are.

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I've been talking
about trees and stuff.

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Here I mean chain, otherwise
there aren't two end points

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to think about.

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Good.

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So here are his
NP-hardness proofs.

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[INAUDIBLE] the problems
are NP-complete.

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They're pretty simple.

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The problem here we're
reducing from is partition.

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I give you a bunch of integers.

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I want to divide them into
two halves of equal sum.

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And the top example is
minimum flat span problem.

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So you make an orthogonal chain
where the horizontal edges are

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long and they're proportional
to the integers you're given,

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the vertical edges
are really tiny.

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And so what you'd like to do--
all you can do is sort of flip

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because you have to
stay in the plane,

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you can flip one of
the vertical edges,

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say, and make any of these
edges go left or right.

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You get that freedom.

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So each integer,
you get to choose.

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Do I go right by that amount and
or do I go left by that amount?

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And if the amount you go left
is equal to the amount you

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go right-- in other
words, is it partitioned

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into two equal sums-- then
those endpoints will be aligned,

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and then their distance
will be very tiny.

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Otherwise, it will
be quite large

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because the horizontal
distances are all big.

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So it's kind of a very
easy NP-hardness proof.

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To maximize your
flat span, instead

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of mapping your
integers on to lengths,

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you map them on to
angles-- return angles.

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I won't specify
that too precisely.

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But again, if you make
your total counterclockwise

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turn equal to your
total clockwise turn,

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then the two end edges,
which are super, super long,

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will be parallel.

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And to maximize the distance
between the endpoints,

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you want them to be parallel.

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If you make them go some
other angle, they're closer.

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Now, both of these proofs
rely on the requirement

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that you want a
flat configuration

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with minimum or maximum span.

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Now, there's a claim
that flat configurations

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matter for proteins, so
it's a natural constraint.

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But what about the
general problem?

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What about, I have something
in three dimensions,

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I want to maximize-- I have
a fixed angle chain in 3D,

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maximize or minimize the span?

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Both of those problems are open.

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Can you solve them
in polynomial time?

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For 3D max span, so the non flat
version just for maximization,

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there's been a lot of work.

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And there are two
papers on the subject.

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One of them is by
Nadia and Joe O'Rourke.

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Another one is by
Borcea and Streinu.

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And I just want to
quickly summarize

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that because there's
a lot of stuff there.

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But essentially, they find what
the structure of those spans

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look like.

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I have an early figure
that's in our book

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before all this work was done.

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The simple chain, this black guy
at 1, 2 3, 4 bars open chain,

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and in that black three
dimensional state,

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it maximizes the span,
the green span there.

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And if you look
from above, which

248
00:12:33,980 --> 00:12:37,400
is this picture-- of course,
the end points look much closer

249
00:12:37,400 --> 00:12:42,800
in projection-- and the red
configuration is the max span

250
00:12:42,800 --> 00:12:45,050
if you restrict to
flat configurations.

251
00:12:45,050 --> 00:12:47,790
So here, of course,
3D buys you something.

252
00:12:47,790 --> 00:12:49,870
In general, it always will.

253
00:12:49,870 --> 00:12:51,710
An interesting thing
is that this max

254
00:12:51,710 --> 00:12:56,740
span-- the green line--
passes through another vertex.

255
00:12:56,740 --> 00:12:58,340
It seems kind of weird.

256
00:12:58,340 --> 00:13:01,740
And in fact, there's a
general theorem there

257
00:13:01,740 --> 00:13:05,762
sort of characterizing the
structure of these chains.

258
00:13:05,762 --> 00:13:07,220
It's still not
known whether we can

259
00:13:07,220 --> 00:13:08,830
solve this problem
in polynomial time.

260
00:13:08,830 --> 00:13:23,820
But for orthogonal chains, where
all the angles are 90 degrees,

261
00:13:23,820 --> 00:13:27,580
we can solve that in
linear time, I guess.

262
00:13:31,490 --> 00:13:33,950
And here's what it looks like.

263
00:13:39,395 --> 00:13:41,020
Suppose you have some
orthogonal chain.

264
00:13:41,020 --> 00:13:43,186
Orthogonal chains are nice
because you can draw them

265
00:13:43,186 --> 00:13:44,890
in the plane as a staircase.

266
00:13:44,890 --> 00:13:46,805
So there's a nice
canonical configuration.

267
00:13:50,420 --> 00:13:56,830
One way to think about
how to find the max span

268
00:13:56,830 --> 00:14:00,040
configuration-- I'm just
going to give a high level

269
00:14:00,040 --> 00:14:02,390
overview here, this won't
be a complete algorithm--

270
00:14:02,390 --> 00:14:05,100
is you triangulated
that staircase

271
00:14:05,100 --> 00:14:07,034
in this sort of obvious
way of connecting

272
00:14:07,034 --> 00:14:08,575
every endpoint to
the one, two ahead.

273
00:14:12,290 --> 00:14:16,280
And think about this as like a
body that's hinging around here

274
00:14:16,280 --> 00:14:19,519
because I can spin-- if I spin
the left part of this chain

275
00:14:19,519 --> 00:14:21,560
around this edge, it's
like hinging that triangle

276
00:14:21,560 --> 00:14:22,780
around that hinge.

277
00:14:22,780 --> 00:14:23,564
Same thing.

278
00:14:23,564 --> 00:14:24,980
You could think
of these triangles

279
00:14:24,980 --> 00:14:27,520
as just being hinged together,
like in rigid origami.

280
00:14:30,070 --> 00:14:32,940
It's the same class of motions.

281
00:14:32,940 --> 00:14:36,100
And now you can--
what I'm going to do

282
00:14:36,100 --> 00:14:41,100
is compute a shortest path in
this surface from here to here.

283
00:14:41,100 --> 00:14:43,440
Confusingly, this is called
a geodesic shortest path

284
00:14:43,440 --> 00:14:45,390
although it's not really
related to geodesics

285
00:14:45,390 --> 00:14:47,870
from polyhedral surfaces.

286
00:14:47,870 --> 00:14:49,490
But if I compute
a shortest path,

287
00:14:49,490 --> 00:14:51,310
it's going to go
like to this vertex

288
00:14:51,310 --> 00:14:53,495
and then probably
to that vertex.

289
00:14:53,495 --> 00:14:56,020
But I'm constrained to
stay inside the union

290
00:14:56,020 --> 00:14:56,850
of those triangles.

291
00:14:56,850 --> 00:14:59,340
I want to go from one
endpoint to another.

292
00:14:59,340 --> 00:15:03,690
Then I claim that--
OK, these two edges

293
00:15:03,690 --> 00:15:07,300
will stay planar, of course
they form a triangle-- I claim

294
00:15:07,300 --> 00:15:10,270
these four edges
will stay planar,

295
00:15:10,270 --> 00:15:12,920
and in the orthogonal
case they'll stay zigzag.

296
00:15:12,920 --> 00:15:16,430
And then also these two guys
will stay in their own plane.

297
00:15:16,430 --> 00:15:19,740
And then I claim that actually
this wiggly line, which

298
00:15:19,740 --> 00:15:22,060
is not straight
because it bends here

299
00:15:22,060 --> 00:15:24,620
and it bends here, the total
length of that wiggly line

300
00:15:24,620 --> 00:15:26,180
is the max span.

301
00:15:26,180 --> 00:15:29,580
And you achieve that by folding
this planar part with respect

302
00:15:29,580 --> 00:15:32,190
to this planar part with
respect to this planar part

303
00:15:32,190 --> 00:15:36,532
so that the wiggly lines
become aligned and straight.

304
00:15:36,532 --> 00:15:37,740
And that's very hard to draw.

305
00:15:37,740 --> 00:15:39,720
But it can be done,
and that's what

306
00:15:39,720 --> 00:15:42,100
you do in the orthogonal
case and that gives you

307
00:15:42,100 --> 00:15:45,760
the answer in linear
time with enough work.

308
00:15:48,400 --> 00:15:50,680
For non orthogonal
though, it's open

309
00:15:50,680 --> 00:15:52,880
whether you can do this
in polynomial time.

310
00:15:52,880 --> 00:15:54,693
Maybe it's NP-hard, actually.

311
00:15:54,693 --> 00:15:55,234
I don't know.

312
00:16:05,960 --> 00:16:08,490
That's all I want
to say about span.

313
00:16:08,490 --> 00:16:09,785
Next, we go to flattening.

314
00:16:20,550 --> 00:16:22,640
The first question
about flattening,

315
00:16:22,640 --> 00:16:24,884
and the main one
we'll talk about here

316
00:16:24,884 --> 00:16:26,550
until we get to flat
state connectivity,

317
00:16:26,550 --> 00:16:32,740
is does a fixed angle chain
have a flat state at all?

318
00:16:32,740 --> 00:16:35,880
Can you even draw it in
the plane without crossing?

319
00:16:35,880 --> 00:16:40,480
So we're restricted here to
have no self intersections.

320
00:16:40,480 --> 00:16:47,180
We want flat state,
no self intersection.

321
00:16:50,812 --> 00:16:53,270
Then there would be a question
of given some configuration,

322
00:16:53,270 --> 00:16:55,660
can I actually continuously
get to a flat state?

323
00:16:55,660 --> 00:16:58,650
But the simplest question
is ignore getting there.

324
00:16:58,650 --> 00:17:01,140
Just, is there a flat state?

325
00:17:01,140 --> 00:17:04,290
And this problem is NP-hard.

326
00:17:04,290 --> 00:17:06,880
Again, Mike Soss and his
advisor Godfried Toussaint.

327
00:17:09,717 --> 00:17:11,800
It's a little more
complicated, but it's basically

328
00:17:11,800 --> 00:17:15,280
the same idea as that very
simple proof, which was just

329
00:17:15,280 --> 00:17:20,400
to map integers to a little
zigzag staircase here.

330
00:17:20,400 --> 00:17:23,554
So the goal is to
force x to end up

331
00:17:23,554 --> 00:17:25,470
being-- the two endpoints
of the green curve--

332
00:17:25,470 --> 00:17:26,480
to be aligned with each other.

333
00:17:26,480 --> 00:17:28,030
That will exist if
and only if there

334
00:17:28,030 --> 00:17:30,850
is a partition of
given integers.

335
00:17:30,850 --> 00:17:33,080
And there's all this
infrastructure that's

336
00:17:33,080 --> 00:17:37,100
sort of-- there's little
lock here and a key,

337
00:17:37,100 --> 00:17:39,470
and some structure on the left.

338
00:17:39,470 --> 00:17:43,990
Basically forces the
picture to look like that.

339
00:17:43,990 --> 00:17:46,200
So the first claim is
that the black stuff

340
00:17:46,200 --> 00:17:47,510
is basically unique.

341
00:17:47,510 --> 00:17:49,150
I think there's one
global reflection

342
00:17:49,150 --> 00:17:51,290
you can do that doesn't
affect anything.

343
00:17:51,290 --> 00:17:53,270
But you try any of
the other flips.

344
00:17:53,270 --> 00:17:55,190
Again, we're restricted
to flat states here.

345
00:17:55,190 --> 00:17:57,470
So there's only sort of a
bounded number of things

346
00:17:57,470 --> 00:17:59,760
you can do, a finite number
of things you can do.

347
00:17:59,760 --> 00:18:01,890
You try all of them,
they self intersect.

348
00:18:01,890 --> 00:18:03,980
So the black thing
is basically forced,

349
00:18:03,980 --> 00:18:06,050
and it forces the
endpoint-- this endpoint

350
00:18:06,050 --> 00:18:08,420
x-- from the black
side to be aligned

351
00:18:08,420 --> 00:18:10,970
with this very narrow spike.

352
00:18:10,970 --> 00:18:12,740
And because the
angles are preserved,

353
00:18:12,740 --> 00:18:14,240
that red guy's going
to be vertical.

354
00:18:14,240 --> 00:18:16,320
It can't go down
so it must go up.

355
00:18:16,320 --> 00:18:19,530
And so only if this thing is
aligned in the center, aligned

356
00:18:19,530 --> 00:18:21,770
with x-- in other
words, this problem

357
00:18:21,770 --> 00:18:26,580
has a partition-- will
this have a flat state.

358
00:18:26,580 --> 00:18:28,890
So it's not the most
exciting example.

359
00:18:28,890 --> 00:18:31,620
This is only a weak
NP-hardness proof.

360
00:18:31,620 --> 00:18:33,450
Lots of interesting
questions still

361
00:18:33,450 --> 00:18:37,950
open here, like if all the
links are the same, if they're

362
00:18:37,950 --> 00:18:41,270
all equal, then we don't know.

363
00:18:41,270 --> 00:18:43,760
Or if all the links are
even polynomially bounded,

364
00:18:43,760 --> 00:18:46,390
this needs really, really long
lengths verses really, really

365
00:18:46,390 --> 00:18:50,640
tiny links exponentially--
exponential in ratio.

366
00:18:50,640 --> 00:18:53,080
All these problems are open.

367
00:18:53,080 --> 00:18:54,680
And that's flattening.

368
00:18:54,680 --> 00:18:58,657
So we're going very quickly
because there isn't-- well,

369
00:18:58,657 --> 00:19:00,490
partly because I'm more
excited about this--

370
00:19:00,490 --> 00:19:03,420
but there's more work
in these two parts.

371
00:19:03,420 --> 00:19:05,220
So I'm going to focus on that.

372
00:19:05,220 --> 00:19:07,890
Next topic is flat
state connectivity.

373
00:19:21,540 --> 00:19:24,870
So the idea is to think
about the configuration

374
00:19:24,870 --> 00:19:28,272
space of these fixed
angle chains, let's say.

375
00:19:28,272 --> 00:19:29,730
And we kind of know
that it's going

376
00:19:29,730 --> 00:19:31,980
to be disconnected because
there are knitting needles,

377
00:19:31,980 --> 00:19:33,320
there are nasty things.

378
00:19:33,320 --> 00:19:36,505
So there's maybe various
connective components.

379
00:19:40,410 --> 00:19:44,475
But let's say that we really
care about flat states.

380
00:19:47,590 --> 00:19:51,080
And the question is, are
they connected to each other?

381
00:19:51,080 --> 00:19:53,350
So in other words, do
all the flat states--

382
00:19:53,350 --> 00:19:55,940
mark them with x's--
do they all appear?

383
00:19:55,940 --> 00:19:57,330
There's only finitely many.

384
00:19:57,330 --> 00:19:59,445
So configurations,
there's this continuum

385
00:19:59,445 --> 00:20:02,420
that there are these messy
blobs, semi algebraic sets.

386
00:20:02,420 --> 00:20:04,885
The flat states, those
are discrete things.

387
00:20:04,885 --> 00:20:06,260
Because we have
fixed angles, you

388
00:20:06,260 --> 00:20:07,737
can flip or not flip every edge.

389
00:20:07,737 --> 00:20:09,320
So [INAUDIBLE] most
exponentially many

390
00:20:09,320 --> 00:20:11,300
of them, so finite.

391
00:20:11,300 --> 00:20:12,930
Are they all in one component?

392
00:20:12,930 --> 00:20:16,400
So I can get-- if I pick two
of my favorite flat states,

393
00:20:16,400 --> 00:20:18,090
there's a path between them?

394
00:20:18,090 --> 00:20:21,560
Or are some of them in
multiple components?

395
00:20:21,560 --> 00:20:23,785
So in this case, we call
it flat state disconnected.

396
00:20:27,730 --> 00:20:32,530
And if they're all like this,
we call it flat state connected.

397
00:20:32,530 --> 00:20:35,420
And we'd just like to
know which chains, which

398
00:20:35,420 --> 00:20:39,050
fixed angle trees, whatever,
are flat state connected

399
00:20:39,050 --> 00:20:41,340
versus flat state disconnected.

400
00:20:41,340 --> 00:20:44,610
I would say, the
big open problem

401
00:20:44,610 --> 00:20:52,110
here is every fixed angle
chain, open chain flat state

402
00:20:52,110 --> 00:20:53,572
connected?

403
00:20:53,572 --> 00:20:54,910
That is still open.

404
00:20:54,910 --> 00:21:01,220
We have lots of results
in that direction.

405
00:21:01,220 --> 00:21:03,620
So the top four results
are about open chains,

406
00:21:03,620 --> 00:21:05,510
but they have an
extra constraint.

407
00:21:05,510 --> 00:21:08,730
For example, open chains that
have a monotone configuration,

408
00:21:08,730 --> 00:21:10,520
like the staircase.

409
00:21:10,520 --> 00:21:14,440
Those are flat state connected.

410
00:21:14,440 --> 00:21:17,210
In fact, whenever the
angles between the edges

411
00:21:17,210 --> 00:21:22,250
are either orthogonal
or obtuse, then they're

412
00:21:22,250 --> 00:21:24,030
flat state connected.

413
00:21:24,030 --> 00:21:26,400
When the angles are acute,
we're not really sure.

414
00:21:26,400 --> 00:21:29,760
If all the angles are equal
and acute, then we can do it.

415
00:21:29,760 --> 00:21:33,090
But if they're different
and acute, we don't know.

416
00:21:33,090 --> 00:21:36,310
Unless the edges
are all unit length

417
00:21:36,310 --> 00:21:39,210
and the angles are in this
funny range, then we can do it.

418
00:21:39,210 --> 00:21:42,020
So there's all these
special cases we can solve.

419
00:21:42,020 --> 00:21:45,770
The most relevant to proteins
is actually obtuse chains,

420
00:21:45,770 --> 00:21:48,470
so we've solved sort
of the main problem

421
00:21:48,470 --> 00:21:50,910
with this second result.

422
00:21:50,910 --> 00:21:52,680
But there's a natural
theoretical question

423
00:21:52,680 --> 00:21:55,970
here is, are all open
chains flat state connected

424
00:21:55,970 --> 00:21:57,590
or do we get disconnectivity?

425
00:21:57,590 --> 00:22:00,100
I will show you that-- I'll
show the orthogonal case

426
00:22:00,100 --> 00:22:01,692
in a little bit.

427
00:22:01,692 --> 00:22:03,900
We can do some stuff if you
have multiple chains that

428
00:22:03,900 --> 00:22:07,100
are attached to some
blob like a cell.

429
00:22:07,100 --> 00:22:13,490
Closed chains is a little
bit-- for disconnected,

430
00:22:13,490 --> 00:22:16,970
we don't have very interesting
examples, I would say.

431
00:22:16,970 --> 00:22:18,770
This is funny because
locked examples

432
00:22:18,770 --> 00:22:22,330
are easy to come by but flat
state disconnected examples are

433
00:22:22,330 --> 00:22:25,550
little trickier because
flat is so constrained.

434
00:22:25,550 --> 00:22:29,320
So let me just show
you these examples.

435
00:22:29,320 --> 00:22:33,740
This is what we call a partially
rigid fixed angle tree.

436
00:22:33,740 --> 00:22:36,510
So not only are
the angles fixed,

437
00:22:36,510 --> 00:22:41,120
but also the black
edges are not-- in fact,

438
00:22:41,120 --> 00:22:43,650
only the blue edges here
are allowed to spin.

439
00:22:43,650 --> 00:22:45,680
Everything else is held rigid.

440
00:22:45,680 --> 00:22:49,310
So these arms are somehow forced
to be in exactly that geometry.

441
00:22:49,310 --> 00:22:52,210
I can spin it
around this edge, so

442
00:22:52,210 --> 00:22:54,580
spin it up into 3D, for example.

443
00:22:54,580 --> 00:22:59,390
These are two different flat
states of the same linkage.

444
00:22:59,390 --> 00:23:02,160
The only difference between
these two-- I haven't rotated

445
00:23:02,160 --> 00:23:05,160
or anything-- is that I've
taken each of these arms

446
00:23:05,160 --> 00:23:08,740
and flipped it
around a blue axis.

447
00:23:08,740 --> 00:23:10,880
If I do all four of them,
I would get this picture.

448
00:23:10,880 --> 00:23:14,570
But the claim is, you cannot do
that without self intersection.

449
00:23:14,570 --> 00:23:17,170
The intuition is, when
there aren't very-- oh,

450
00:23:17,170 --> 00:23:20,152
one other thing that makes
it slightly more interesting.

451
00:23:20,152 --> 00:23:22,610
It's weird to say, well why
did you force some of the edges

452
00:23:22,610 --> 00:23:24,940
to be rigid and not others?

453
00:23:24,940 --> 00:23:28,150
One way to force that is
to use a general graph.

454
00:23:28,150 --> 00:23:30,350
If you add some extra
edges to sort of brace this

455
00:23:30,350 --> 00:23:32,750
and all these angles are
fixed, then this linkage

456
00:23:32,750 --> 00:23:35,340
will behave exactly
like that one.

457
00:23:35,340 --> 00:23:38,560
So that at least is
somewhat more natural,

458
00:23:38,560 --> 00:23:43,448
although what we really care
about are chains, maybe trees.

459
00:23:43,448 --> 00:23:44,822
But we don't know
whether there's

460
00:23:44,822 --> 00:23:48,450
a-- we also don't know
whether all fixed angle

461
00:23:48,450 --> 00:23:50,470
trees are flat state connected.

462
00:23:50,470 --> 00:23:54,100
These are the worst
examples we know.

463
00:23:54,100 --> 00:23:57,920
Let me give you an idea
of why it doesn't work.

464
00:23:57,920 --> 00:24:01,940
This is a little animation
of just a couple of moves

465
00:24:01,940 --> 00:24:06,690
attempted, and it's just
going cycle through that.

466
00:24:06,690 --> 00:24:12,910
And these are some static images
of the same kind of thing.

467
00:24:12,910 --> 00:24:15,480
So the intuition is the
following-- you have four arms.

468
00:24:15,480 --> 00:24:19,250
You have two sides to the
plane, there's up and down.

469
00:24:19,250 --> 00:24:23,000
The four arms and two
sides, at least two of them

470
00:24:23,000 --> 00:24:24,780
are going to have to
go to the same side.

471
00:24:24,780 --> 00:24:27,020
The best you can do is two
and two, or three and one.

472
00:24:27,020 --> 00:24:29,640
But in either case,
you have two sides

473
00:24:29,640 --> 00:24:32,700
go to the-- two arms
that go on the same side.

474
00:24:32,700 --> 00:24:35,400
Now, it could be,
like in this image,

475
00:24:35,400 --> 00:24:37,100
that they're opposite arms.

476
00:24:37,100 --> 00:24:41,260
So there's this arm here
and there's this arm here.

477
00:24:41,260 --> 00:24:44,260
So they're connected
by a 180 degree angle.

478
00:24:44,260 --> 00:24:46,460
And those guys,
when they fold up,

479
00:24:46,460 --> 00:24:50,160
actually these edges will
just hit each other dead on.

480
00:24:50,160 --> 00:24:53,395
So that's kind of obvious
from a geometric standpoint.

481
00:24:53,395 --> 00:24:55,520
Maybe you call it cheating
for them to hit dead on.

482
00:24:55,520 --> 00:24:57,978
You can twiddle the edge lengths
so that they will properly

483
00:24:57,978 --> 00:25:01,060
intersect without
dead on collision,

484
00:25:01,060 --> 00:25:04,040
without being
degenerate basically.

485
00:25:04,040 --> 00:25:05,624
The alternative
is that-- and this

486
00:25:05,624 --> 00:25:07,290
is a little harder
to see geometrically,

487
00:25:07,290 --> 00:25:13,450
and that's why we drew that
animation-- is that you have

488
00:25:13,450 --> 00:25:16,340
one arm and you have an adjacent
arm connected by a 90 degree

489
00:25:16,340 --> 00:25:17,410
angle.

490
00:25:17,410 --> 00:25:19,560
Now here, there's clearly
some collision going on.

491
00:25:19,560 --> 00:25:21,890
And if you happen to fold
it up 90 degrees like that

492
00:25:21,890 --> 00:25:24,170
and then fold the other guy,
obviously you get stuck.

493
00:25:24,170 --> 00:25:26,090
But maybe you fold it a
little bit and the other guy

494
00:25:26,090 --> 00:25:27,790
goes a little bit
more and there could

495
00:25:27,790 --> 00:25:30,650
be some dance between those two
degrees of freedom, those two

496
00:25:30,650 --> 00:25:34,410
arms, that somehow gets
them both to pass over

497
00:25:34,410 --> 00:25:35,290
to the other side.

498
00:25:35,290 --> 00:25:36,456
It's obviously not possible.

499
00:25:36,456 --> 00:25:37,830
How do you prove it?

500
00:25:37,830 --> 00:25:41,900
Well, you could prove it with
topology-- knot theory or link

501
00:25:41,900 --> 00:25:43,330
theory.

502
00:25:43,330 --> 00:25:45,390
So it's a very cute proof.

503
00:25:45,390 --> 00:25:48,650
You start with-- so
here's the full example,

504
00:25:48,650 --> 00:25:51,010
but I've highlighted
the two arms in red

505
00:25:51,010 --> 00:25:52,990
that are going to move.

506
00:25:52,990 --> 00:25:55,660
And I imagine
connecting the endpoints

507
00:25:55,660 --> 00:25:59,720
of each arm with these
little blue ropes

508
00:25:59,720 --> 00:26:01,870
underneath the plane.

509
00:26:01,870 --> 00:26:03,370
They're both going
on the same side.

510
00:26:03,370 --> 00:26:05,286
Let's say they somehow
pass through each other

511
00:26:05,286 --> 00:26:06,310
on the top side.

512
00:26:06,310 --> 00:26:08,220
Then I'm free to connect
stuff on the bottom,

513
00:26:08,220 --> 00:26:10,400
and I shouldn't
collide with that.

514
00:26:10,400 --> 00:26:12,930
So if somehow,
both of these guys

515
00:26:12,930 --> 00:26:18,350
flip over-- so arm on
the left, A3 flips over.

516
00:26:18,350 --> 00:26:21,080
A3 stays where it
is but now the arm

517
00:26:21,080 --> 00:26:24,580
is on the top, the north side
instead of the south side.

518
00:26:24,580 --> 00:26:30,490
And the other guy, from B
to B3, used to go like this

519
00:26:30,490 --> 00:26:31,800
and now it goes like this.

520
00:26:31,800 --> 00:26:34,560
If that happens somehow,
then these ropes

521
00:26:34,560 --> 00:26:36,450
could remain intact
during that whole motion.

522
00:26:36,450 --> 00:26:41,184
On the top, you have two closed
loops that are not interlocked.

523
00:26:41,184 --> 00:26:42,850
On the bottom, you
have two closed loops

524
00:26:42,850 --> 00:26:43,747
that are interlocked.

525
00:26:43,747 --> 00:26:45,580
So there's no way to
get from there to there

526
00:26:45,580 --> 00:26:47,050
without colliding somewhere.

527
00:26:47,050 --> 00:26:48,970
The blue stuff didn't
move, so the red stuff

528
00:26:48,970 --> 00:26:50,610
must have collided.

529
00:26:50,610 --> 00:26:52,770
So even just topologically,
you are screwed.

530
00:26:56,310 --> 00:26:59,130
That is their only
negative example.

531
00:26:59,130 --> 00:27:02,420
Lots of interesting
open questions here.

532
00:27:02,420 --> 00:27:06,690
On the positive
side, let me show you

533
00:27:06,690 --> 00:27:09,210
for orthogonal chains--
and the same algorithm

534
00:27:09,210 --> 00:27:11,300
works for obtuse
chains, all the angles

535
00:27:11,300 --> 00:27:15,310
are obtuse-- how they
are flat state connected.

536
00:27:15,310 --> 00:27:17,910
So in order to show it's
flat state connected,

537
00:27:17,910 --> 00:27:20,260
I want to think
about two flat states

538
00:27:20,260 --> 00:27:22,500
and show that I
can fold from one

539
00:27:22,500 --> 00:27:27,620
to the other via some
intermediate 3D stuff.

540
00:27:27,620 --> 00:27:30,650
Let's start with one
of the flat states.

541
00:27:30,650 --> 00:27:32,260
So it's orthogonal.

542
00:27:32,260 --> 00:27:34,959
So in two dimensions,
all the edges

543
00:27:34,959 --> 00:27:36,250
will be horizontal or vertical.

544
00:27:36,250 --> 00:27:39,929
In 3D, they can kind of be in
many, many different angles,

545
00:27:39,929 --> 00:27:41,345
many different
dihedral triangles.

546
00:27:41,345 --> 00:27:43,500
In 2D, it's pretty simple.

547
00:27:43,500 --> 00:27:46,006
And all I need to do is
sort of pick up that chain,

548
00:27:46,006 --> 00:27:48,130
and I'm going to try to
pick it up into a staircase

549
00:27:48,130 --> 00:27:49,750
because there's
only one staircase.

550
00:27:49,750 --> 00:27:52,690
If I can make it a
staircase, I make

551
00:27:52,690 --> 00:27:55,010
flat configuration A a
staircase, flat configuration B

552
00:27:55,010 --> 00:27:58,230
a staircase, and just
FedEx in the middle.

553
00:27:58,230 --> 00:28:00,300
Once they're both
staircases, I play one motion

554
00:28:00,300 --> 00:28:03,279
and the other one backwards,
get from anywhere to anywhere.

555
00:28:03,279 --> 00:28:04,320
So here's all you do you.

556
00:28:04,320 --> 00:28:06,170
You take the first
edge and you just

557
00:28:06,170 --> 00:28:13,126
rotate it up to the red line A.
And then you take the next edge

558
00:28:13,126 --> 00:28:14,500
and you take both
of those edges,

559
00:28:14,500 --> 00:28:16,930
and you just rotate
them like this,

560
00:28:16,930 --> 00:28:21,130
so you get that little
2-step staircase.

561
00:28:21,130 --> 00:28:22,950
Now I'd really like
to pick up this edge,

562
00:28:22,950 --> 00:28:24,750
but I want to
first get these two

563
00:28:24,750 --> 00:28:26,940
edges in a plane with that edge.

564
00:28:26,940 --> 00:28:32,740
So I rotate this flag over to
the left, I get those two guys.

565
00:28:32,740 --> 00:28:35,750
And now they're in a plane with
this, and I just lift that up.

566
00:28:35,750 --> 00:28:38,200
Then I'm going to
flip, then rotate up.

567
00:28:38,200 --> 00:28:40,600
Flip, rotate, flip, rotate.

568
00:28:40,600 --> 00:28:42,810
Here's some more examples.

569
00:28:42,810 --> 00:28:48,150
So if at this point, I have
this staircase-- sorry,

570
00:28:48,150 --> 00:28:52,650
I guess originally I have
from V3 to D up there.

571
00:28:52,650 --> 00:28:56,310
it's not in plane with this guy,
so I just rotate it like that.

572
00:28:56,310 --> 00:28:58,530
I'm spinning around this edge.

573
00:28:58,530 --> 00:29:01,790
So now I have from
B3 to #, and then I

574
00:29:01,790 --> 00:29:04,510
rotate it up along
that green arc.

575
00:29:04,510 --> 00:29:07,785
And I get a bigger
staircase above the chain

576
00:29:07,785 --> 00:29:09,410
and because everything's
staying above,

577
00:29:09,410 --> 00:29:10,950
it will never
penetrate the plane

578
00:29:10,950 --> 00:29:12,690
and will never hit anybody else.

579
00:29:12,690 --> 00:29:14,740
And I'm building a
staircase by design.

580
00:29:14,740 --> 00:29:16,550
I always rotate this--
there's actually

581
00:29:16,550 --> 00:29:18,500
two ways I could be in
plane-- but I always

582
00:29:18,500 --> 00:29:20,400
rotate it so that when
I pick an edge up,

583
00:29:20,400 --> 00:29:22,280
it'll be in a staircase.

584
00:29:22,280 --> 00:29:23,940
So this is actually really easy.

585
00:29:23,940 --> 00:29:27,310
And slight generalization
is to obtuse chains,

586
00:29:27,310 --> 00:29:34,336
then instead of making a
staircase, we make a monotone.

587
00:29:34,336 --> 00:29:35,455
Let me get this right.

588
00:29:38,580 --> 00:29:40,440
Yeah, sum z monotone state.

589
00:29:40,440 --> 00:29:43,550
So it goes monotone and z,
out of the plane, that's

590
00:29:43,550 --> 00:29:45,400
enough to avoid
collision, and you

591
00:29:45,400 --> 00:29:47,100
get a canonical configuration.

592
00:29:47,100 --> 00:29:51,620
Also, if you have acute angles
but all the angles are equal,

593
00:29:51,620 --> 00:29:54,240
then there's a
natural conical state,

594
00:29:54,240 --> 00:29:56,320
which is just like a
compressed staircase.

595
00:29:56,320 --> 00:29:57,830
And that will work here, too.

596
00:29:57,830 --> 00:29:58,902
That takes more effort.

597
00:29:58,902 --> 00:30:00,110
That was in a separate paper.

598
00:30:03,000 --> 00:30:06,140
But big open question is,
chains with arbitrary angles.

599
00:30:06,140 --> 00:30:07,620
We have no idea.

600
00:30:07,620 --> 00:30:11,176
It is very hard to do
an operation like this.

601
00:30:11,176 --> 00:30:12,550
Wow, we are burning
through this.

602
00:30:12,550 --> 00:30:13,140
This is fun.

603
00:30:15,670 --> 00:30:22,875
So the next topic is
about locked chains.

604
00:30:22,875 --> 00:30:24,860
Now as I said, you can
take a knitting needles

605
00:30:24,860 --> 00:30:28,580
example, which has five edges.

606
00:30:34,011 --> 00:30:35,510
And that will still
be locked if you

607
00:30:35,510 --> 00:30:37,840
force the angles to be
fixed because it was locked

608
00:30:37,840 --> 00:30:39,910
without the angles being fixed.

609
00:30:39,910 --> 00:30:44,540
Now, it required a length
ratio of 3:1, I think.

610
00:30:44,540 --> 00:30:46,980
This edge had to be longer
than the sum of those three.

611
00:30:50,440 --> 00:30:55,230
So let me put down
some open problems.

612
00:30:55,230 --> 00:31:00,180
So you may recall in the
case of universal chains--

613
00:31:00,180 --> 00:31:12,660
universal joints, I should
say-- the big open question was,

614
00:31:12,660 --> 00:31:16,250
can you lock a
universal joint 3D chain

615
00:31:16,250 --> 00:31:19,860
with unit edge lengths?

616
00:31:19,860 --> 00:31:24,900
So, equilateral-- every
edge is the same length.

617
00:31:24,900 --> 00:31:27,170
Is there a locked chain
like the knitting needles

618
00:31:27,170 --> 00:31:28,795
when all the edge
lengths are the same.

619
00:31:28,795 --> 00:31:31,420
And one of the motivations
for that is in proteins,

620
00:31:31,420 --> 00:31:35,390
the edge lengths are all
within like 50% of each other.

621
00:31:35,390 --> 00:31:38,900
So it's pretty
natural, of course.

622
00:31:38,900 --> 00:31:44,316
We don't have universal
joints with proteins.

623
00:31:44,316 --> 00:31:45,440
We have fixed angle joints.

624
00:31:48,560 --> 00:31:52,670
So the big open problem
for fixed angle joints--

625
00:31:52,670 --> 00:32:05,660
I guess we'll do this in parts--
is there a locked 3D fixed

626
00:32:05,660 --> 00:32:24,600
angle chain that's equilateral?

627
00:32:24,600 --> 00:32:26,200
I'm going to add
some conditions here.

628
00:32:30,910 --> 00:32:32,560
So that's the first
natural question.

629
00:32:32,560 --> 00:32:34,780
Knitting needles
doesn't suffice.

630
00:32:34,780 --> 00:32:36,900
We need a 3:1 length
ratio, as far as we know.

631
00:32:43,100 --> 00:32:45,180
Turns out that question's
not very interesting.

632
00:32:45,180 --> 00:32:48,600
I need to do slightly
nonlinear editing here.

633
00:32:48,600 --> 00:32:50,990
So you take your
knitting needles example,

634
00:32:50,990 --> 00:32:53,920
and you just subdivide the edges
into lots of little tiny bars.

635
00:32:53,920 --> 00:32:55,500
It doesn't have to
be this extreme.

636
00:32:55,500 --> 00:32:57,660
You could not subdivide
these edges at all,

637
00:32:57,660 --> 00:32:59,350
and make these
guys subdivide them

638
00:32:59,350 --> 00:33:01,350
into like three or four parts.

639
00:33:01,350 --> 00:33:04,310
Because the angles
are fixed, these guys

640
00:33:04,310 --> 00:33:06,360
act as a single [INAUDIBLE].

641
00:33:06,360 --> 00:33:08,010
There's really no difference.

642
00:33:08,010 --> 00:33:09,920
Maybe you make a
slight curve there

643
00:33:09,920 --> 00:33:14,100
and then they can bend a little
bit, but really not much.

644
00:33:14,100 --> 00:33:17,070
So if you just say, oh, I
want it to be unit length.

645
00:33:17,070 --> 00:33:20,720
I don't constrain what the
angles are but I fix them,

646
00:33:20,720 --> 00:33:23,570
then it's trivial to come
up with locked examples.

647
00:33:23,570 --> 00:33:27,000
So that's not very interesting.

648
00:33:27,000 --> 00:33:29,380
What if I make it not
only equilateral--

649
00:33:29,380 --> 00:33:31,740
the lengths are the same--
if I make it equiangular.

650
00:33:37,850 --> 00:33:41,750
Because, again, in proteins,
all the angles are similar.

651
00:33:41,750 --> 00:33:45,150
They're around 110, 108,
something like that.

652
00:33:45,150 --> 00:33:48,750
They're all pretty close,
I think within 10 to 20%

653
00:33:48,750 --> 00:33:50,450
of each other.

654
00:33:50,450 --> 00:33:52,580
Well, there's also
a locked example.

655
00:33:52,580 --> 00:33:56,120
And just to show you how
research was done back

656
00:33:56,120 --> 00:34:00,790
at the turn of the century,
this is pre-web 2.0, pre-Ajax

657
00:34:00,790 --> 00:34:02,380
and all that fancy stuff.

658
00:34:02,380 --> 00:34:04,800
We used Ascii Art.

659
00:34:04,800 --> 00:34:07,050
Email was the tool of choice.

660
00:34:07,050 --> 00:34:12,780
I know it's hard to imagine
a time-- 2002, so long ago.

661
00:34:12,780 --> 00:34:14,110
And I tracked this down.

662
00:34:14,110 --> 00:34:16,889
This is the original
claim it looks--

663
00:34:16,889 --> 00:34:18,630
we call this the
crossed legs example

664
00:34:18,630 --> 00:34:21,300
because it's like two legs
crossed around each other.

665
00:34:21,300 --> 00:34:24,760
And this is the first time
we thought, oh, maybe it

666
00:34:24,760 --> 00:34:26,210
could be done, unit length.

667
00:34:26,210 --> 00:34:29,300
This is Stefan Langerman.

668
00:34:29,300 --> 00:34:32,400
And here, for the
first time ever--

669
00:34:32,400 --> 00:34:35,270
this is not the first model,
but this is the first photograph

670
00:34:35,270 --> 00:34:39,420
of any model I'm aware of-- this
is the crossed legs example.

671
00:34:39,420 --> 00:34:41,590
This is made with a
construction toy that

672
00:34:41,590 --> 00:34:44,860
used to be sold around here
but is no longer in production.

673
00:34:44,860 --> 00:34:46,290
So they're pretty hard to get.

674
00:34:46,290 --> 00:34:50,090
It's straws-- nicely colored
straws-- and the cool part

675
00:34:50,090 --> 00:34:52,830
are these connectors.

676
00:34:52,830 --> 00:34:55,239
So the connectors force
particular angles.

677
00:34:55,239 --> 00:34:57,800
In this case, every
angle is 45 degrees.

678
00:34:57,800 --> 00:35:00,260
So this is equiangular
and equilateral

679
00:35:00,260 --> 00:35:02,560
because all the straws, I'm
told, are the same length.

680
00:35:02,560 --> 00:35:04,410
That's how they're sold.

681
00:35:04,410 --> 00:35:07,850
And you can do edge spins.

682
00:35:07,850 --> 00:35:11,640
Whoops, that's called cheating.

683
00:35:11,640 --> 00:35:15,400
It's not totally obvious
that this is locked.

684
00:35:15,400 --> 00:35:18,680
The problem with the model
is that the edges can bend.

685
00:35:18,680 --> 00:35:26,060
But if you treat it properly
and only spin around the edges,

686
00:35:26,060 --> 00:35:28,840
then you're stuck.

687
00:35:28,840 --> 00:35:32,260
Now, there is one
thing you can do.

688
00:35:32,260 --> 00:35:39,600
See if I-- yeah, like this.

689
00:35:39,600 --> 00:35:41,140
So here, I'm almost in a plane.

690
00:35:41,140 --> 00:35:44,820
I've got the purple edge
right against the pink one.

691
00:35:44,820 --> 00:35:46,070
Easier to see from that angle?

692
00:35:46,070 --> 00:35:47,370
I don't know.

693
00:35:47,370 --> 00:35:51,230
So here, this guy can
come out and this guy

694
00:35:51,230 --> 00:35:54,162
can barely go on the edge.

695
00:35:54,162 --> 00:35:56,370
So actually, this doesn't
quite work for equilateral.

696
00:35:56,370 --> 00:35:57,720
It works for one plus epsilon.

697
00:35:57,720 --> 00:36:01,420
That's why I added these
little nubs at the end.

698
00:36:01,420 --> 00:36:04,090
So if they're all
exactly equal length

699
00:36:04,090 --> 00:36:07,930
and you allow just
abrasion of the endpoint,

700
00:36:07,930 --> 00:36:09,550
then this could go
around like that

701
00:36:09,550 --> 00:36:11,950
and then you'd be unlocked.

702
00:36:11,950 --> 00:36:14,090
But if you just add
slightly-- either

703
00:36:14,090 --> 00:36:16,310
you change the angles
to be not quite equal,

704
00:36:16,310 --> 00:36:18,770
so make this a little smaller,
or you make the lengths

705
00:36:18,770 --> 00:36:21,159
a little bit longer
at the ends-- then

706
00:36:21,159 --> 00:36:22,200
the claim is it's locked.

707
00:36:22,200 --> 00:36:24,116
We don't actually have
a formal proof of this.

708
00:36:24,116 --> 00:36:26,750
We're just remembering, hey, we
should probably write this up.

709
00:36:26,750 --> 00:36:28,208
I was talking to
Stefan last night.

710
00:36:32,060 --> 00:36:34,260
So someday we'll prove
that this is locked.

711
00:36:34,260 --> 00:36:35,760
But it certainly looks like it.

712
00:36:35,760 --> 00:36:38,540
So this isn't open yet.

713
00:36:38,540 --> 00:36:41,180
I mean, modulo the
details of that proof.

714
00:36:41,180 --> 00:36:43,700
Equilateral and
equiangular seems

715
00:36:43,700 --> 00:36:46,370
easy to lock with
fixed angle chains.

716
00:36:46,370 --> 00:36:51,270
In fact, even easier, this
example only has four edges.

717
00:36:51,270 --> 00:36:54,060
So even less than
the knitting needles.

718
00:36:54,060 --> 00:37:01,410
Fixed angles make for
complicated motions, I guess.

719
00:37:01,410 --> 00:37:03,250
Make it hard to unlock things.

720
00:37:03,250 --> 00:37:05,140
So I need to add
one more constraint,

721
00:37:05,140 --> 00:37:08,780
and the constraint is obtuse.

722
00:37:08,780 --> 00:37:11,650
So again, all of
these properties

723
00:37:11,650 --> 00:37:15,170
are enjoyed by proteins.

724
00:37:15,170 --> 00:37:17,520
Protein backbones have
all these properties.

725
00:37:17,520 --> 00:37:20,430
Even if you looked
at fixed angle trees,

726
00:37:20,430 --> 00:37:22,610
is there something like
this that's locked?

727
00:37:22,610 --> 00:37:25,070
And now, we don't know.

728
00:37:25,070 --> 00:37:26,790
And this seems quite tricky.

729
00:37:26,790 --> 00:37:29,319
I guess the intuition is that
obtuse-- and usually we think

730
00:37:29,319 --> 00:37:30,985
about orthogonal,
just cause it's easier

731
00:37:30,985 --> 00:37:32,480
to draw the
pictures, but reality

732
00:37:32,480 --> 00:37:37,010
is more like 108 degrees--
the conjecture is

733
00:37:37,010 --> 00:37:41,062
obtuse fixed angle chains behave
kind of like universal joints.

734
00:37:41,062 --> 00:37:42,520
And with universal
joints, we don't

735
00:37:42,520 --> 00:37:45,370
know whether
equilateral is enough.

736
00:37:45,370 --> 00:37:49,026
So it's tricky.

737
00:37:49,026 --> 00:37:51,224
Yeah, question.

738
00:37:51,224 --> 00:37:52,640
AUDIENCE: In the
previous example,

739
00:37:52,640 --> 00:37:57,697
you showed the ribbon thing--

740
00:37:57,697 --> 00:37:58,780
PROFESSOR: The subdivided.

741
00:37:58,780 --> 00:38:01,238
AUDIENCE: Subdivided into a
bunch of little interconnecting

742
00:38:01,238 --> 00:38:02,098
pieces.

743
00:38:02,098 --> 00:38:07,000
What if you, instead,
made your ribbon lengths

744
00:38:07,000 --> 00:38:13,365
basically a bunch of little
unit obtuse angle connectors,

745
00:38:13,365 --> 00:38:16,850
and then when you hit
the big terms it's just

746
00:38:16,850 --> 00:38:18,540
obtuse, obtuse, obtuse, obtuse.

747
00:38:18,540 --> 00:38:20,630
PROFESSOR: Yeah, you
can make this example

748
00:38:20,630 --> 00:38:21,630
be entirely obtuse.

749
00:38:21,630 --> 00:38:23,040
You can make every angle obtuse.

750
00:38:23,040 --> 00:38:24,750
Here, you could
arc a little bit.

751
00:38:24,750 --> 00:38:27,550
Here, you could arc some
more, but not too sharp.

752
00:38:27,550 --> 00:38:29,200
And because here,
we actually know

753
00:38:29,200 --> 00:38:30,880
that this part can
be made a string.

754
00:38:30,880 --> 00:38:32,596
We don't really care
what it looks like.

755
00:38:32,596 --> 00:38:33,970
So you can make
it fairly obtuse.

756
00:38:33,970 --> 00:38:36,000
It's just that these guys
should not bend much.

757
00:38:36,000 --> 00:38:39,560
They have to be long no
matter how you fold them.

758
00:38:39,560 --> 00:38:45,890
So if you want equilateral
and obtuse, that's also easy.

759
00:38:45,890 --> 00:38:48,660
But to make all the
angles actually be equal,

760
00:38:48,660 --> 00:38:51,260
as far as we know you cannot
take that knitting needles

761
00:38:51,260 --> 00:38:52,202
subdivided.

762
00:38:52,202 --> 00:38:54,410
Make all the lengths equal,
and all the angles equal,

763
00:38:54,410 --> 00:38:56,370
and make them obtuse.

764
00:38:56,370 --> 00:38:57,300
That's open.

765
00:38:57,300 --> 00:38:59,040
But any two out of
the three, it's easy.

766
00:39:02,650 --> 00:39:05,090
Of course, in reality they're
not quite equilateral.

767
00:39:05,090 --> 00:39:06,690
They're not quite equiangular.

768
00:39:06,690 --> 00:39:07,940
But it's still open for those.

769
00:39:07,940 --> 00:39:09,900
If you have like a small
range for the lengths

770
00:39:09,900 --> 00:39:12,170
and a small range for
the angles, this is open.

771
00:39:12,170 --> 00:39:15,600
We pose it this way because
it's the cleanest geometrically.

772
00:39:15,600 --> 00:39:17,560
But the real question
you care about

773
00:39:17,560 --> 00:39:21,750
is when these are
fuzzy constraints.

774
00:39:21,750 --> 00:39:26,790
Obtuse is real, but
these guys are fuzzier.

775
00:39:26,790 --> 00:39:32,980
So if you think about proteins,
which fold very well in nature,

776
00:39:32,980 --> 00:39:36,190
there are a couple of
reasons they might fold well.

777
00:39:36,190 --> 00:39:39,150
We know, as far as
fixed angle chains go,

778
00:39:39,150 --> 00:39:41,650
it's actually quite easy
to find locked examples.

779
00:39:41,650 --> 00:39:45,089
And, this is somewhat
intuitive but bear with me,

780
00:39:45,089 --> 00:39:47,630
because they are locked examples
in this configuration space,

781
00:39:47,630 --> 00:39:51,860
we believe these configuration
spaces are really ugly nasty.

782
00:39:51,860 --> 00:39:55,430
So it would be very hard--
even if you know, oh I only

783
00:39:55,430 --> 00:39:58,560
need to fold something
in my component--

784
00:39:58,560 --> 00:40:00,120
if these guys are
highly disconnected

785
00:40:00,120 --> 00:40:02,040
and flat states are
all over the place,

786
00:40:02,040 --> 00:40:05,000
it's probably even within
this connected component,

787
00:40:05,000 --> 00:40:06,740
it looks really ugly.

788
00:40:06,740 --> 00:40:10,650
And so it's very hard to find a
path from one state to another.

789
00:40:10,650 --> 00:40:12,690
Probably pieced based
complete, although we

790
00:40:12,690 --> 00:40:15,840
don't know that for sure.

791
00:40:15,840 --> 00:40:16,940
But that's the intuition.

792
00:40:16,940 --> 00:40:19,384
Locked equals messy.

793
00:40:19,384 --> 00:40:21,050
When there are no
locked configurations,

794
00:40:21,050 --> 00:40:23,800
like carpenter's rules, we
get really nice algorithms.

795
00:40:23,800 --> 00:40:27,620
It's super easy to get from
state A to state B. Now,

796
00:40:27,620 --> 00:40:29,790
if you're nature or
you're designing nature,

797
00:40:29,790 --> 00:40:33,660
let's say, or you're building
your own virtual world, Second

798
00:40:33,660 --> 00:40:36,409
Life, and you want
to design proteins,

799
00:40:36,409 --> 00:40:38,200
you would like to design
them in such a way

800
00:40:38,200 --> 00:40:40,740
that they fold easily because
it happens all the time.

801
00:40:40,740 --> 00:40:44,500
Every thing that is being acted
on by our body, every living

802
00:40:44,500 --> 00:40:46,910
thing that we know has
tons of little proteins

803
00:40:46,910 --> 00:40:48,120
that are doing all the work.

804
00:40:48,120 --> 00:40:51,340
They are folded into their
shape and they do something.

805
00:40:51,340 --> 00:40:54,570
That's proteins plus
RNA, but mostly proteins.

806
00:40:54,570 --> 00:40:58,480
So to understand life, we
should understand proteins.

807
00:40:58,480 --> 00:41:02,000
Now, how to proteins
fold so well when we know

808
00:41:02,000 --> 00:41:04,480
there are all these
locked configurations?

809
00:41:04,480 --> 00:41:07,010
One possible answer is that
proteins have extra structure,

810
00:41:07,010 --> 00:41:09,240
namely these three
things, which somehow

811
00:41:09,240 --> 00:41:11,120
make it very easy to
algorithmically go

812
00:41:11,120 --> 00:41:16,370
from A to B. Notice I'm not
assuming anything about how

813
00:41:16,370 --> 00:41:21,450
proteins fold in terms of what
is the mechanism that drives

814
00:41:21,450 --> 00:41:25,760
them because we don't really
understand those mechanisms.

815
00:41:25,760 --> 00:41:28,730
There's hydrophobia, which we
don't really know how it works.

816
00:41:28,730 --> 00:41:35,052
So all these little forces
that we don't fully understand.

817
00:41:35,052 --> 00:41:36,760
We understand lots of
parts of the story,

818
00:41:36,760 --> 00:41:39,096
but not the whole story.

819
00:41:39,096 --> 00:41:41,220
And what's convenient about
these kinds of problems

820
00:41:41,220 --> 00:41:42,719
is you don't need
to assume anything

821
00:41:42,719 --> 00:41:44,746
about how it actually happens.

822
00:41:44,746 --> 00:41:46,620
All we're assuming is
the mechanical behavior

823
00:41:46,620 --> 00:41:50,140
of the proteins, and how
they could possibly fold.

824
00:41:50,140 --> 00:41:53,780
And the idea is if there's
locked configurations, that's

825
00:41:53,780 --> 00:41:56,530
probably the wrong model
because then everything's messy.

826
00:41:56,530 --> 00:41:58,420
Now there's also
evolution coming into play

827
00:41:58,420 --> 00:42:00,128
and maybe some proteins
are easy to fold,

828
00:42:00,128 --> 00:42:01,861
some proteins are hard to fold.

829
00:42:01,861 --> 00:42:03,360
That's an interesting
question which

830
00:42:03,360 --> 00:42:05,540
should be experimented with.

831
00:42:05,540 --> 00:42:08,782
But let's hope that
there's a model.

832
00:42:08,782 --> 00:42:09,990
Things are mutating randomly.

833
00:42:09,990 --> 00:42:12,570
You really like
everything to fold nicely.

834
00:42:12,570 --> 00:42:14,500
Maybe it's because
you have all three

835
00:42:14,500 --> 00:42:16,420
of these properties,
approximately,

836
00:42:16,420 --> 00:42:17,200
in real proteins.

837
00:42:19,950 --> 00:42:24,270
So general idea is that nature
has some extra constraints that

838
00:42:24,270 --> 00:42:27,180
make protein folding easy.

839
00:42:27,180 --> 00:42:29,510
Just have to figure
out what they are

840
00:42:29,510 --> 00:42:30,790
and why it makes them easy.

841
00:42:30,790 --> 00:42:32,190
Unfortunately, this is
still an open problem.

842
00:42:32,190 --> 00:42:34,690
If this had an algorithm, that
would be a natural candidate

843
00:42:34,690 --> 00:42:37,400
for what nature's doing
using its mechanical--

844
00:42:37,400 --> 00:42:42,420
or using it's energies
and forces, and so on.

845
00:42:42,420 --> 00:42:44,450
This would be a rather
unsatisfactory ending

846
00:42:44,450 --> 00:42:49,570
if this was-- if the
climax was an open problem.

847
00:42:49,570 --> 00:42:50,640
We have a theorem too.

848
00:42:53,150 --> 00:42:55,615
And this is what I'll
cover in most detail.

849
00:43:00,350 --> 00:43:04,620
And it's a paper called
producible protein chains.

850
00:43:04,620 --> 00:43:09,590
Protein chains just means fixed
angle chains, open chains.

851
00:43:09,590 --> 00:43:13,680
And the idea is well, yeah,
there are these constraints

852
00:43:13,680 --> 00:43:15,160
or there are these
extra features.

853
00:43:15,160 --> 00:43:16,576
We don't know how
to exploit them,

854
00:43:16,576 --> 00:43:18,300
so let's not even
worry about them.

855
00:43:18,300 --> 00:43:20,410
Suppose they don't even exist.

856
00:43:20,410 --> 00:43:23,525
Maybe I'm going to assume
obtuse, but none of the others.

857
00:43:26,220 --> 00:43:30,030
There's another
constraint in how proteins

858
00:43:30,030 --> 00:43:32,495
fold, or really how
proteins are created.

859
00:43:35,650 --> 00:43:38,990
They're created by a machine,
a molecular machine made up

860
00:43:38,990 --> 00:43:43,632
of a whole bunch of proteins
and RNA, called the ribosome.

861
00:43:43,632 --> 00:43:44,980
You may have heard of.

862
00:43:44,980 --> 00:43:48,520
It translates messenger
RNA into proteins.

863
00:43:48,520 --> 00:43:51,300
So there's some mRNA
around here, maybe.

864
00:43:51,300 --> 00:43:53,300
Don't know exactly how
this machine works.

865
00:43:53,300 --> 00:43:56,120
But there are actually very
accurate three dimensional

866
00:43:56,120 --> 00:43:58,370
reconstructions of the ribosome.

867
00:43:58,370 --> 00:43:59,970
With no copyright
free images, you're

868
00:43:59,970 --> 00:44:02,094
going to have to-- there's
a link on the slide that

869
00:44:02,094 --> 00:44:07,240
goes to the cool and 3D
models of the ribosome,

870
00:44:07,240 --> 00:44:08,320
with a slice away.

871
00:44:08,320 --> 00:44:11,010
So you can see there's
a tunnel down here,

872
00:44:11,010 --> 00:44:13,740
and the protein get
sort of created here.

873
00:44:13,740 --> 00:44:15,690
The background
gets created here.

874
00:44:15,690 --> 00:44:17,550
And it starts going
through this tunnel.

875
00:44:17,550 --> 00:44:19,120
There's a bend in
the tunnel around

876
00:44:19,120 --> 00:44:23,340
here, where it's conjectured
and an amino acid gets attached.

877
00:44:23,340 --> 00:44:25,910
And then it goes out the
tunnel and the protein

878
00:44:25,910 --> 00:44:28,460
starts spewing out here
and presumably folding

879
00:44:28,460 --> 00:44:29,180
at the same time.

880
00:44:29,180 --> 00:44:31,770
We don't really know.

881
00:44:31,770 --> 00:44:34,020
So this is how
proteins are created.

882
00:44:36,620 --> 00:44:39,030
The birds and bees,
I guess, of proteins.

883
00:44:39,030 --> 00:44:42,670
So what's interesting
about this is

884
00:44:42,670 --> 00:44:45,790
it's not like a protein
exists and then folds,

885
00:44:45,790 --> 00:44:48,120
which is how a lot
of people might

886
00:44:48,120 --> 00:44:50,499
think about it at first glance.

887
00:44:50,499 --> 00:44:52,540
That's the natural way to
model protein folding--

888
00:44:52,540 --> 00:44:55,270
start with a protein, say,
and just zigzag configuration.

889
00:44:55,270 --> 00:44:58,420
If it's obtuse, there's a nice
zigzag monotone configuration.

890
00:44:58,420 --> 00:45:00,750
Then you see what is
the best configuration

891
00:45:00,750 --> 00:45:02,580
I could fold into, for
some notion of best.

892
00:45:05,500 --> 00:45:07,500
And that's sort of what
this configuration space

893
00:45:07,500 --> 00:45:08,450
picture is about.

894
00:45:08,450 --> 00:45:10,950
It's if I already
have a protein, what

895
00:45:10,950 --> 00:45:13,000
configurations can
I reach by motions?

896
00:45:13,000 --> 00:45:14,000
And that is interesting.

897
00:45:14,000 --> 00:45:15,541
That's important
because you're still

898
00:45:15,541 --> 00:45:17,050
going to have to
reach by a motion.

899
00:45:17,050 --> 00:45:20,210
But, it's actually
more flexible than that

900
00:45:20,210 --> 00:45:23,720
because the protein could
just be partially built.

901
00:45:23,720 --> 00:45:26,720
The rest of the protein
hasn't been built.

902
00:45:26,720 --> 00:45:28,882
And it could start
folding already.

903
00:45:28,882 --> 00:45:30,340
It might be easier
to fold when you

904
00:45:30,340 --> 00:45:33,460
don't have the obstacles
of your existing protein.

905
00:45:33,460 --> 00:45:39,050
So that's both a worry, but
it's also a convenient structure

906
00:45:39,050 --> 00:45:43,160
because this ribosome
is a giant obstacle.

907
00:45:43,160 --> 00:45:45,550
Bigger than most proteins.

908
00:45:45,550 --> 00:45:48,110
If your protein's really long,
maybe it could go over here.

909
00:45:48,110 --> 00:45:52,300
But most the time, it's going to
stay on one side of this plane

910
00:45:52,300 --> 00:45:55,220
because locally, this thing
is basically flat, if you look

911
00:45:55,220 --> 00:45:59,330
at the real 3D pictures,
not the schematic.

912
00:45:59,330 --> 00:46:02,020
Now, this is good news for a
geometer because there's this

913
00:46:02,020 --> 00:46:05,340
giant obstacle-- think
of it as a half space--

914
00:46:05,340 --> 00:46:08,210
which the protein cannot
penetrate while it's being

915
00:46:08,210 --> 00:46:09,050
produced over here.

916
00:46:12,820 --> 00:46:14,010
That's it.

917
00:46:14,010 --> 00:46:18,030
That half space constraint
is enough to get

918
00:46:18,030 --> 00:46:21,100
really good algorithms
for folding your chain.

919
00:46:21,100 --> 00:46:24,140
It's weird because we've
made a problem both harder

920
00:46:24,140 --> 00:46:27,350
because the protein is only
partially produced at any time

921
00:46:27,350 --> 00:46:29,272
and it can fold,
which is part of it,

922
00:46:29,272 --> 00:46:30,730
but we've also made
our life easier

923
00:46:30,730 --> 00:46:32,900
because there's
this big obstacle.

924
00:46:32,900 --> 00:46:34,500
AUDIENCE: [INAUDIBLE]
makes sense

925
00:46:34,500 --> 00:46:37,710
why there's only obtuse
angles there, right?

926
00:46:37,710 --> 00:46:38,669
PROFESSOR: Yeah, right.

927
00:46:38,669 --> 00:46:41,293
Out of this, we're going to get
that the angles and the protein

928
00:46:41,293 --> 00:46:42,310
are constrained.

929
00:46:42,310 --> 00:46:48,430
And in particular, for
this angle-- it depends.

930
00:46:48,430 --> 00:46:54,536
I mean, in this picture because
it's perpendicular here, yeah,

931
00:46:54,536 --> 00:46:55,910
the sharpest angle
you could make

932
00:46:55,910 --> 00:46:58,870
is 90 degrees, more or less.

933
00:46:58,870 --> 00:47:00,760
That's a good point.

934
00:47:00,760 --> 00:47:04,390
So it's a convenient match
between the chemistry, which

935
00:47:04,390 --> 00:47:07,300
also forces the angles
to be obtuse, I guess.

936
00:47:07,300 --> 00:47:09,030
I don't know a ton of chemistry.

937
00:47:09,030 --> 00:47:11,380
But also, the ribosome
just geometrically forces.

938
00:47:11,380 --> 00:47:13,240
We're going to use a
property like that.

939
00:47:13,240 --> 00:47:17,150
Our model is going to be a
little bit more-- both more

940
00:47:17,150 --> 00:47:20,190
general and simpler.

941
00:47:20,190 --> 00:47:23,860
We're going to imagine that
the ribosome is a cone.

942
00:47:23,860 --> 00:47:25,800
It's part of the
upper come here.

943
00:47:25,800 --> 00:47:29,320
This is like a mirror image.

944
00:47:29,320 --> 00:47:32,400
And in reality, that
cone is actually

945
00:47:32,400 --> 00:47:34,202
a plane and everything
above the plane.

946
00:47:34,202 --> 00:47:35,660
But to be more
general, we're going

947
00:47:35,660 --> 00:47:37,944
to allow some angle alpha here.

948
00:47:37,944 --> 00:47:40,360
It's also just easier to think
about when alpha is smaller

949
00:47:40,360 --> 00:47:43,860
than 90, but everything I say
will work when alpha equals 90

950
00:47:43,860 --> 00:47:47,120
and that is sort of
the reality case.

951
00:47:47,120 --> 00:47:52,560
So the model is-- so
the ribosome is a cone.

952
00:47:52,560 --> 00:47:54,680
We call this the half
angle of the cone.

953
00:47:54,680 --> 00:47:58,511
From the vertical axis to the
edge of the cone is alpha.

954
00:47:58,511 --> 00:48:00,510
So if you were going from
one axis to the other,

955
00:48:00,510 --> 00:48:03,240
it would be 2 alpha.

956
00:48:03,240 --> 00:48:08,610
The model is, you start with
one link of your chain, which

957
00:48:08,610 --> 00:48:09,920
is inside the cone.

958
00:48:09,920 --> 00:48:14,960
It's spews out through the apex.

959
00:48:14,960 --> 00:48:16,850
That's the exit of the tunnel.

960
00:48:16,850 --> 00:48:19,930
Here, we're allowing the tunnel
to be actually quite free.

961
00:48:19,930 --> 00:48:23,920
Doesn't have to be
perpendicular to the apex

962
00:48:23,920 --> 00:48:25,860
or the plane of the apex.

963
00:48:25,860 --> 00:48:27,920
So the edge comes
out, and as soon

964
00:48:27,920 --> 00:48:31,850
as the endpoint of the
chain reaches here,

965
00:48:31,850 --> 00:48:33,430
then a new link is created.

966
00:48:33,430 --> 00:48:38,150
This is like a very simple model
for how a chain can come out

967
00:48:38,150 --> 00:48:40,050
of a cone without
worrying about what's

968
00:48:40,050 --> 00:48:41,940
happening inside the cone.

969
00:48:41,940 --> 00:48:44,277
Imagining everything's
totally free.

970
00:48:44,277 --> 00:48:46,110
This is like you can
allow self intersection

971
00:48:46,110 --> 00:48:47,472
in the cone, who knows what.

972
00:48:47,472 --> 00:48:48,930
But once you come
outside the cone,

973
00:48:48,930 --> 00:48:50,570
you're not allowed
to self intersect

974
00:48:50,570 --> 00:48:53,120
and you're not allowed
to intersect the cone.

975
00:48:53,120 --> 00:48:56,560
Once you come out,
you can't go back in.

976
00:48:56,560 --> 00:49:00,860
So that is a model of
producing protein chains.

977
00:49:00,860 --> 00:49:04,240
And if you have a
cone of angle alpha,

978
00:49:04,240 --> 00:49:07,270
we call this an alpha
producible chain.

979
00:49:10,090 --> 00:49:13,962
For whatever reason, we often
call it beta producible chain.

980
00:49:13,962 --> 00:49:15,260
Just change the variable.

981
00:49:40,660 --> 00:49:43,930
So if you think of the ribosome
as a cone with half angle beta,

982
00:49:43,930 --> 00:49:45,610
you can produce it like this.

983
00:49:45,610 --> 00:49:46,610
That is beta producible.

984
00:49:46,610 --> 00:49:48,540
Now this is a pretty
powerful model

985
00:49:48,540 --> 00:49:51,175
because you only have to
worry about it link by link.

986
00:49:51,175 --> 00:49:53,300
You don't have to worry
about the rest of the chain

987
00:49:53,300 --> 00:49:55,492
until it spews
outside of the cone.

988
00:49:55,492 --> 00:49:57,950
But it's restrictive in that
you cannot penetrate the cone.

989
00:50:01,560 --> 00:50:02,820
All right.

990
00:50:02,820 --> 00:50:06,810
One thing we can
talk about is angles.

991
00:50:06,810 --> 00:50:11,720
So I'm going to
write call a chain

992
00:50:11,720 --> 00:50:14,810
a less than or
equal to alpha chain

993
00:50:14,810 --> 00:50:18,005
if all the turn angles are
less than or equal to alpha.

994
00:50:22,420 --> 00:50:24,860
I don't know if I've used
turn angles in this class.

995
00:50:24,860 --> 00:50:26,900
Probably.

996
00:50:26,900 --> 00:50:31,270
If I have two edges,
the angle would be this.

997
00:50:31,270 --> 00:50:33,400
The turn angle would be
this, the supplement.

998
00:50:38,110 --> 00:50:38,610
Yeah.

999
00:50:38,610 --> 00:50:40,110
I guess we used
turn angles way back

1000
00:50:40,110 --> 00:50:42,381
in origami land, Kawasaki's
theorem and so on.

1001
00:50:42,381 --> 00:50:43,880
It's just, if you're
going straight,

1002
00:50:43,880 --> 00:50:47,750
how much do you have to turn
to get to the next edge.

1003
00:50:47,750 --> 00:50:50,430
So we'd like fairly
obtuse things.

1004
00:50:50,430 --> 00:50:51,880
So alpha is going to be small.

1005
00:50:51,880 --> 00:50:53,760
There isn't a ton of turn.

1006
00:50:53,760 --> 00:50:56,870
But in general, less than or
equal to alpha chain for some

1007
00:50:56,870 --> 00:50:57,440
alpha.

1008
00:50:57,440 --> 00:50:59,570
Now there's a relation--
as Jason was mentioning,

1009
00:50:59,570 --> 00:51:03,150
there's a relation between
alpha and beta in the ribosome

1010
00:51:03,150 --> 00:51:04,970
because you always
exited orthogonally

1011
00:51:04,970 --> 00:51:08,730
to the plane that was your cone.

1012
00:51:08,730 --> 00:51:14,120
The sharpest angle you could
get was 90 degree turn angle.

1013
00:51:14,120 --> 00:51:18,170
Here, we're a little freer
because this edge can wiggle

1014
00:51:18,170 --> 00:51:20,950
around as long as
it touches the apex.

1015
00:51:20,950 --> 00:51:24,237
So if you're up
against the cone,

1016
00:51:24,237 --> 00:51:26,570
you have to slide out into
the complimentary cone-- that

1017
00:51:26,570 --> 00:51:31,490
was the previous picture--
and as soon as you get there,

1018
00:51:31,490 --> 00:51:34,310
you could create a new
edge which is like this.

1019
00:51:34,310 --> 00:51:39,110
So the sharpest angle you can
get is actually twice beta.

1020
00:51:39,110 --> 00:51:44,932
In general, we're going to
have alpha over 2 is less than

1021
00:51:44,932 --> 00:51:45,640
or equal to beta.

1022
00:51:45,640 --> 00:51:50,670
That is-- you can get up
to beta equals 2 alpha.

1023
00:51:50,670 --> 00:51:51,420
Get that right.

1024
00:51:56,940 --> 00:52:01,300
And also in the obtuse case,
this is not too exciting.

1025
00:52:01,300 --> 00:52:01,980
But it's true.

1026
00:52:05,640 --> 00:52:07,140
There's actually
some problems here.

1027
00:52:07,140 --> 00:52:08,990
When you have that
full flexibility

1028
00:52:08,990 --> 00:52:15,760
and you set alpha to two beta,
not the other way around.

1029
00:52:15,760 --> 00:52:22,430
I'm going to assume here
that alpha equals beta.

1030
00:52:22,430 --> 00:52:24,500
This will be convenient.

1031
00:52:24,500 --> 00:52:28,080
And it's the interesting case
because, in reality, the cone

1032
00:52:28,080 --> 00:52:29,860
has a half angle of 90 degrees.

1033
00:52:29,860 --> 00:52:32,110
So beta is 90.

1034
00:52:32,110 --> 00:52:33,890
And the sharpest angle
we're going to make

1035
00:52:33,890 --> 00:52:35,020
was always obtuse.

1036
00:52:35,020 --> 00:52:38,080
So saying that you have a
less than or equal to 90 chain

1037
00:52:38,080 --> 00:52:39,510
is just fine.

1038
00:52:39,510 --> 00:52:41,670
But on the mathematical
side, I think

1039
00:52:41,670 --> 00:52:46,010
we saw the case when alpha is
less than or equal to beta,

1040
00:52:46,010 --> 00:52:49,280
but not when alpha over 2 is
less than or equal to beta.

1041
00:52:49,280 --> 00:52:51,520
That's a weaker constraint.

1042
00:52:51,520 --> 00:52:54,380
So there is a range
where it's not so easy.

1043
00:52:58,350 --> 00:53:01,270
Now, what do I claim
about these chains other

1044
00:53:01,270 --> 00:53:05,240
than their angles
are not so sharp?

1045
00:53:05,240 --> 00:53:10,580
I claim they're good
algorithms for folding them.

1046
00:53:10,580 --> 00:53:12,450
What could I possibly mean?

1047
00:53:12,450 --> 00:53:14,690
There are still
locked configurations.

1048
00:53:14,690 --> 00:53:16,560
Is that true?

1049
00:53:16,560 --> 00:53:22,010
Well, I mean presumably--
this is acute--

1050
00:53:22,010 --> 00:53:25,267
but you take the obtuse
versions of this guy.

1051
00:53:25,267 --> 00:53:27,600
Because I didn't constrain
the edge lengths or anything,

1052
00:53:27,600 --> 00:53:30,640
I just said that the
angles are obtuse.

1053
00:53:30,640 --> 00:53:33,540
So I could just sort of round
these corners, make it obtuse.

1054
00:53:33,540 --> 00:53:36,050
You know, add lots of
dots just at the corners

1055
00:53:36,050 --> 00:53:37,360
that would be obtuse.

1056
00:53:37,360 --> 00:53:40,900
And a chain like this
will be producible.

1057
00:53:40,900 --> 00:53:43,190
A chain with these angles
and these edge lengths

1058
00:53:43,190 --> 00:53:45,040
can be produced from a cone.

1059
00:53:45,040 --> 00:53:49,700
But this configuration of
this chain cannot be produced,

1060
00:53:49,700 --> 00:53:50,760
I claim.

1061
00:53:50,760 --> 00:53:53,760
I claim anything
that can be produced

1062
00:53:53,760 --> 00:53:56,600
is in one connected component.

1063
00:53:56,600 --> 00:53:58,815
So while I can make a
linkage that is locked

1064
00:53:58,815 --> 00:54:00,940
and that there are bad
configurations you can't get

1065
00:54:00,940 --> 00:54:03,080
out of, the things
you can actually make,

1066
00:54:03,080 --> 00:54:05,220
you can always get out of.

1067
00:54:05,220 --> 00:54:09,980
So there's going to be the space
of producible configurations.

1068
00:54:13,367 --> 00:54:15,200
Maybe there's some stuff
that's unproducible

1069
00:54:15,200 --> 00:54:16,800
but still connected to it.

1070
00:54:16,800 --> 00:54:17,425
I don't know.

1071
00:54:17,425 --> 00:54:18,550
It doesn't matter too much.

1072
00:54:18,550 --> 00:54:20,460
I won't worry about this stuff.

1073
00:54:20,460 --> 00:54:24,450
There's other bad locked
configurations that cannot

1074
00:54:24,450 --> 00:54:25,330
reach here.

1075
00:54:25,330 --> 00:54:28,000
But everything that's producible
is in one connected component

1076
00:54:28,000 --> 00:54:29,510
of the configuration space.

1077
00:54:29,510 --> 00:54:30,800
That's property one.

1078
00:54:30,800 --> 00:54:32,420
That's kind of nice.

1079
00:54:32,420 --> 00:54:36,912
Also, all the flat states
are going to be in here.

1080
00:54:36,912 --> 00:54:38,120
This is actually pretty easy.

1081
00:54:38,120 --> 00:54:40,570
I just need to prove that
flat states are producible,

1082
00:54:40,570 --> 00:54:42,320
which we'll worry about later.

1083
00:54:42,320 --> 00:54:45,930
So in particular, these guys
are flat state connected.

1084
00:54:45,930 --> 00:54:47,740
All the producible
protein chains

1085
00:54:47,740 --> 00:54:49,024
are flat state connected.

1086
00:54:49,024 --> 00:54:50,690
That's interesting
because we don't even

1087
00:54:50,690 --> 00:54:53,060
know that all chains are
flat state connected.

1088
00:54:53,060 --> 00:54:55,730
But here-- I guess we know
that obtuse chains are

1089
00:54:55,730 --> 00:55:01,350
flat state connected, so
maybe it's not so surprising.

1090
00:55:01,350 --> 00:55:03,940
But what's important is not only
are the flat states connected

1091
00:55:03,940 --> 00:55:05,520
to each other and
the producible states

1092
00:55:05,520 --> 00:55:07,311
are connected to each
other, but producible

1093
00:55:07,311 --> 00:55:09,170
is connected to flat states.

1094
00:55:09,170 --> 00:55:10,400
Everything is together here.

1095
00:55:12,912 --> 00:55:14,120
I might have more properties.

1096
00:55:14,120 --> 00:55:15,536
But that's already
some good news.

1097
00:55:15,536 --> 00:55:19,160
And there's algorithms
to do all of this.

1098
00:55:19,160 --> 00:55:21,890
How do we prove it?

1099
00:55:21,890 --> 00:55:31,890
Well, as usual we
use the FedEx method.

1100
00:55:31,890 --> 00:55:34,690
And in some sense,
one of the challenges

1101
00:55:34,690 --> 00:55:46,050
is what is the natural canonical
state for protein chains?

1102
00:55:46,050 --> 00:55:50,596
In fact, we're just going to
assume that our chain is--

1103
00:55:50,596 --> 00:55:51,970
in reality, we're
going to assume

1104
00:55:51,970 --> 00:55:56,080
that it's an orthogonal chain--
an obtuse chain, I should say.

1105
00:55:56,080 --> 00:55:59,040
But in general, for any
less than or equal to alpha

1106
00:55:59,040 --> 00:56:01,300
chain, for whatever
alpha you like-- and it

1107
00:56:01,300 --> 00:56:07,200
will be the half angle of the
cone, so alpha equals beta--

1108
00:56:07,200 --> 00:56:09,135
we will define a
canonical configuration.

1109
00:56:22,010 --> 00:56:23,765
I think we called
it the alpha CCC.

1110
00:56:28,060 --> 00:56:32,440
So it's going to be
kind of like a helix.

1111
00:56:32,440 --> 00:56:36,656
I think I have an example,
an actual computed example.

1112
00:56:36,656 --> 00:56:38,280
That's not the best
picture because you

1113
00:56:38,280 --> 00:56:39,821
can't see everything
that's going on.

1114
00:56:39,821 --> 00:56:42,510
But this is an actual
canonical configuration

1115
00:56:42,510 --> 00:56:44,080
of a particular chain.

1116
00:56:47,980 --> 00:56:50,410
Let me tell you how
it works in general.

1117
00:56:50,410 --> 00:56:54,990
So in general, we have
some chain, v1, v2-- sorry,

1118
00:56:54,990 --> 00:56:58,560
starting at v0, v1, v2.

1119
00:56:58,560 --> 00:57:00,470
I want to define a
canonical-- and there's

1120
00:57:00,470 --> 00:57:02,310
defined lengths between the two.

1121
00:57:02,310 --> 00:57:05,540
And there's defined angles
between every triple

1122
00:57:05,540 --> 00:57:08,240
in sequence.

1123
00:57:08,240 --> 00:57:11,079
So I'm going to start
with v0 somewhere.

1124
00:57:11,079 --> 00:57:12,620
Doesn't really matter
by translation.

1125
00:57:12,620 --> 00:57:15,870
Say, the origin of space.

1126
00:57:15,870 --> 00:57:27,460
And what I'm going to do is draw
a cone whose apex is that v0.

1127
00:57:27,460 --> 00:57:33,410
And the half angle here is
going to be alpha over two,

1128
00:57:33,410 --> 00:57:34,710
not alpha.

1129
00:57:34,710 --> 00:57:38,000
This is a smaller cone
than-- by a factor of two--

1130
00:57:38,000 --> 00:57:40,160
than the ribosome.

1131
00:57:40,160 --> 00:57:42,370
That's important.

1132
00:57:42,370 --> 00:57:45,730
So there's this vertical line.

1133
00:57:45,730 --> 00:57:50,080
And to pick v1, I'm just going
to use the right edge, which

1134
00:57:50,080 --> 00:57:52,960
is-- let's say this
is the x direction.

1135
00:57:52,960 --> 00:57:54,610
This is the z direction.

1136
00:57:54,610 --> 00:57:56,694
Maximum x-coordinate.

1137
00:57:56,694 --> 00:57:58,860
It's going to lie on the
cone, maximum x-coordinate.

1138
00:57:58,860 --> 00:58:00,420
That's v1.

1139
00:58:00,420 --> 00:58:05,560
Now, v1-- let me redraw
this picture a little lower.

1140
00:58:05,560 --> 00:58:10,640
So there was v0, v1.

1141
00:58:10,640 --> 00:58:12,780
Now I want to draw v2, and
I want to draw it above.

1142
00:58:15,390 --> 00:58:20,960
So what I'll do is
draw a vertical column

1143
00:58:20,960 --> 00:58:24,300
whose half angle here
is alpha over two.

1144
00:58:27,270 --> 00:58:29,290
I want to draw v2
on the cone here.

1145
00:58:29,290 --> 00:58:30,780
Of course, the
height of the cone

1146
00:58:30,780 --> 00:58:35,200
is the length of the
edge, not the height.

1147
00:58:35,200 --> 00:58:36,870
You could think of
the cone as infinite.

1148
00:58:36,870 --> 00:58:40,240
And then I just clip this to
when it has the right length.

1149
00:58:40,240 --> 00:58:42,970
So again, this might be
a different height cone.

1150
00:58:42,970 --> 00:58:45,470
I clip it to whatever
the length v1, v2 is.

1151
00:58:45,470 --> 00:58:48,040
I want it to be
somewhere on this cone,

1152
00:58:48,040 --> 00:58:51,070
but now I'm constrained to
have the correct angle at v1.

1153
00:58:51,070 --> 00:58:53,550
I can't just put it over here,
because then the angle here

1154
00:58:53,550 --> 00:58:54,430
would be 180.

1155
00:58:54,430 --> 00:58:57,630
Presumably, I don't want
to make a 180 degree angle.

1156
00:58:57,630 --> 00:58:59,570
So in reality, what
happens is that there's

1157
00:58:59,570 --> 00:59:09,580
a cone which is-- whose
axis is the edge v0, v1.

1158
00:59:09,580 --> 00:59:13,400
So I extend v0 v1 out
here, which, in this case,

1159
00:59:13,400 --> 00:59:15,410
happens to lie here.

1160
00:59:15,410 --> 00:59:18,819
And I make a cone like that.

1161
00:59:18,819 --> 00:59:20,485
Let me draw it slightly
more accurately.

1162
00:59:26,810 --> 00:59:32,550
In this case, the
center axis of the cone

1163
00:59:32,550 --> 00:59:37,990
would go right here, whatever
the extension of v0, v1 was.

1164
00:59:37,990 --> 00:59:45,210
And to have the right angle at
v1, v2 must be on that cone.

1165
00:59:45,210 --> 00:59:47,510
Conveniently, there are
two intersection points

1166
00:59:47,510 --> 00:59:49,300
between those two cones.

1167
00:59:49,300 --> 00:59:52,240
I could choose either
one of them to be v2.

1168
00:59:52,240 --> 00:59:56,300
And I will choose the
counterclockwise most one,

1169
00:59:56,300 --> 00:59:58,560
which is this one.

1170
00:59:58,560 --> 01:00:00,692
This is going to be v2.

1171
01:00:00,692 --> 01:00:03,880
So I draw that edge.

1172
01:00:03,880 --> 01:00:06,090
Now I repeat.

1173
01:00:06,090 --> 01:00:09,130
So for v2, I'm going to
draw a vertical cone whose

1174
01:00:09,130 --> 01:00:12,940
half angle-- here, the half
angle is always alpha over two.

1175
01:00:12,940 --> 01:00:14,930
Half angle is alpha over two.

1176
01:00:14,930 --> 01:00:26,070
This cone had half angle,
whatever the angle v0, v1, v2

1177
01:00:26,070 --> 01:00:27,840
was.

1178
01:00:27,840 --> 01:00:29,460
Was that the half
angle or angle?

1179
01:00:29,460 --> 01:00:30,504
The half angle.

1180
01:00:30,504 --> 01:00:31,420
I think this is right.

1181
01:00:34,080 --> 01:00:36,110
OK.

1182
01:00:36,110 --> 01:00:38,910
And then I take the
intersection of those two cones,

1183
01:00:38,910 --> 01:00:41,750
and that will give
me where v3 is.

1184
01:00:41,750 --> 01:00:44,976
So I do the same thing
for v2, for v3, and so on.

1185
01:00:44,976 --> 01:00:46,600
I have a unique choice
at every moment.

1186
01:00:50,400 --> 01:00:52,890
Yeah, basically.

1187
01:00:52,890 --> 01:00:54,810
And the only exception
was at the beginning

1188
01:00:54,810 --> 01:00:57,340
here, when I had
a vertical line.

1189
01:00:57,340 --> 01:00:59,300
These two cones could
actually be equal,

1190
01:00:59,300 --> 01:01:01,766
and then the intersection
is the entire cone.

1191
01:01:01,766 --> 01:01:03,140
And that case, I
guess I'd choose

1192
01:01:03,140 --> 01:01:05,380
the maximum
x-coordinate one again.

1193
01:01:05,380 --> 01:01:07,350
And then I have a
canonical choice

1194
01:01:07,350 --> 01:01:09,080
of everything along the way.

1195
01:01:09,080 --> 01:01:09,969
It will always go up.

1196
01:01:09,969 --> 01:01:11,510
And it sort of
spinals around because

1197
01:01:11,510 --> 01:01:14,180
of the counter
clockwise most choice.

1198
01:01:14,180 --> 01:01:17,390
And the result is a
picture like this.

1199
01:01:22,446 --> 01:01:23,820
Anything else I
need to say here?

1200
01:01:26,380 --> 01:01:31,190
I claim this canonical
configuration

1201
01:01:31,190 --> 01:01:37,550
lies in an alpha over
2 half angle cone.

1202
01:01:41,270 --> 01:01:44,090
That's true by construction.

1203
01:01:44,090 --> 01:01:47,410
The challenge is, does the
construction really work?

1204
01:01:47,410 --> 01:01:49,445
So I start, obviously,
with one cone,

1205
01:01:49,445 --> 01:01:51,820
and I can think of this as
actually an infinite cone that

1206
01:01:51,820 --> 01:01:53,920
goes out to infinity here.

1207
01:01:53,920 --> 01:01:55,700
And I claim the
entire construction

1208
01:01:55,700 --> 01:01:58,090
will lie inside that cone.

1209
01:01:58,090 --> 01:02:00,930
And it's kind of
obvious because v1--

1210
01:02:00,930 --> 01:02:07,620
I chose v2 to lie in the
same cone, just translated up

1211
01:02:07,620 --> 01:02:10,790
to start at v1 instead
of starting at v0.

1212
01:02:10,790 --> 01:02:13,880
Of course, this cone is
contained in this bigger one.

1213
01:02:13,880 --> 01:02:16,470
And by induction, in fact,
the entire rest of the chain

1214
01:02:16,470 --> 01:02:18,730
will lie in this smaller cone.

1215
01:02:18,730 --> 01:02:22,440
Therefore, it lies
in the big one, also.

1216
01:02:22,440 --> 01:02:24,530
OK.

1217
01:02:24,530 --> 01:02:25,540
Fine.

1218
01:02:25,540 --> 01:02:28,630
So by construction it will
lie in alpha over two cone.

1219
01:02:28,630 --> 01:02:30,825
The worry is that these
two cones don't intersect.

1220
01:02:33,380 --> 01:02:35,100
Here we have an angle.

1221
01:02:35,100 --> 01:02:37,040
The half angle of
the cone is whatever

1222
01:02:37,040 --> 01:02:39,680
angle the angle is at v1.

1223
01:02:39,680 --> 01:02:41,180
Sorry, this should
not be the angle.

1224
01:02:41,180 --> 01:02:42,430
This should be the turn angle.

1225
01:02:48,740 --> 01:02:52,310
That angle is how much
you turn from v0, v1.

1226
01:02:52,310 --> 01:02:54,630
Now, we know that-- we're
assuming that-- the turn

1227
01:02:54,630 --> 01:02:56,010
angles are all, at most, alpha.

1228
01:02:58,700 --> 01:03:01,040
The turn angle
cone could actually

1229
01:03:01,040 --> 01:03:04,620
be twice as big as the vertical
cone that we were always using.

1230
01:03:04,620 --> 01:03:08,640
We always use a vertical cone,
half angle alpha over two.

1231
01:03:08,640 --> 01:03:12,940
But it's OK because we
always keep these edges,

1232
01:03:12,940 --> 01:03:17,750
like v0, v1, was on
the edge of the cone.

1233
01:03:17,750 --> 01:03:19,680
And so when we
extend it, it lies

1234
01:03:19,680 --> 01:03:22,470
on the edge of
this vertical cone.

1235
01:03:22,470 --> 01:03:27,640
So its angle-- in the most
extreme case-- its half angle

1236
01:03:27,640 --> 01:03:31,280
is alpha, which
would look like this.

1237
01:03:31,280 --> 01:03:33,655
We'll go all the way over from
the right side of the cone

1238
01:03:33,655 --> 01:03:34,860
to the left side of the cone.

1239
01:03:34,860 --> 01:03:36,860
In general, it's not going
to be right and left,

1240
01:03:36,860 --> 01:03:40,210
but it's going to be some
side and the antipodal point.

1241
01:03:40,210 --> 01:03:44,610
And because the double
angle of the cone is alpha,

1242
01:03:44,610 --> 01:03:45,700
it's still OK.

1243
01:03:45,700 --> 01:03:49,600
You will intersect
somewhere on the cone.

1244
01:03:49,600 --> 01:03:52,190
This is a subtle detail,
but it's really crucial

1245
01:03:52,190 --> 01:03:54,050
because we start
with a chain that

1246
01:03:54,050 --> 01:03:56,840
has relatively
large angles, alpha.

1247
01:03:56,840 --> 01:03:58,600
And we get it
into-- we squeeze it

1248
01:03:58,600 --> 01:04:01,820
into-- a cone that
still has double-- twice

1249
01:04:01,820 --> 01:04:04,240
its angle is alpha,
but we kind of

1250
01:04:04,240 --> 01:04:08,027
compress it into something
of half angle alpha over two.

1251
01:04:08,027 --> 01:04:10,235
You might think, oh, I'm
just changing the definition

1252
01:04:10,235 --> 01:04:11,454
and calling it half angle.

1253
01:04:11,454 --> 01:04:13,120
Therefore, it gets
to an alpha over two.

1254
01:04:13,120 --> 01:04:15,210
But it's a little
tricky to actually get

1255
01:04:15,210 --> 01:04:18,520
it to fit in a vertical
alpha over two cone.

1256
01:04:18,520 --> 01:04:20,730
Once we have this,
it's really easy

1257
01:04:20,730 --> 01:04:23,470
to canonicalize a chain,
a producible chain.

1258
01:04:23,470 --> 01:04:25,760
So let me tell you
how to do that.

1259
01:04:52,380 --> 01:04:54,510
So we're going use
the FedEx method

1260
01:04:54,510 --> 01:04:57,570
of taking some configuration
and canonicalizing it, and then

1261
01:04:57,570 --> 01:04:59,910
uncanonicalizing
to something else.

1262
01:04:59,910 --> 01:05:02,300
We're also going to use a
new method, which I just

1263
01:05:02,300 --> 01:05:03,360
came up with the term.

1264
01:05:03,360 --> 01:05:05,950
Is called the
momento method which

1265
01:05:05,950 --> 01:05:10,370
is you play the
movie in reverse.

1266
01:05:10,370 --> 01:05:14,130
So, I guess, also
the Merlin method.

1267
01:05:14,130 --> 01:05:16,820
That's more complicated.

1268
01:05:16,820 --> 01:05:21,670
So we have a movie
here in mind, which

1269
01:05:21,670 --> 01:05:24,210
is how was the chain produced?

1270
01:05:24,210 --> 01:05:26,840
So what I want to
show is that if-- I

1271
01:05:26,840 --> 01:05:32,280
want to start with a producible
configuration and chain.

1272
01:05:34,790 --> 01:05:37,180
Somehow, it got produced.

1273
01:05:37,180 --> 01:05:41,440
So you had your
cone and the thing

1274
01:05:41,440 --> 01:05:45,160
starts spewing out and
folding, and doing whatever.

1275
01:05:45,160 --> 01:05:48,807
That's an animation, in some
sense, of one edge coming out

1276
01:05:48,807 --> 01:05:50,390
and stuff is folding
at the same time.

1277
01:05:50,390 --> 01:05:52,139
Then an edge is created,
then another edge

1278
01:05:52,139 --> 01:05:53,250
comes out, and so on.

1279
01:05:53,250 --> 01:05:55,570
What I want to do is play
that movie backwards.

1280
01:05:55,570 --> 01:05:57,690
It's a pretty intuitive idea.

1281
01:05:57,690 --> 01:06:01,310
I just want to start feeding
the edges back into the cone,

1282
01:06:01,310 --> 01:06:03,000
and just keep stuffing them in.

1283
01:06:03,000 --> 01:06:05,060
Now, what happens
out here is easy

1284
01:06:05,060 --> 01:06:07,560
because we know it doesn't
penetrate the cone.

1285
01:06:07,560 --> 01:06:08,960
That's the assumption.

1286
01:06:08,960 --> 01:06:12,120
And we know whatever was
created here could be uncreated,

1287
01:06:12,120 --> 01:06:16,850
as long as you can afford
to erase edges one by one.

1288
01:06:16,850 --> 01:06:18,242
That's the tricky part.

1289
01:06:18,242 --> 01:06:19,200
How do I erase an edge?

1290
01:06:19,200 --> 01:06:21,590
And usually, I can't.

1291
01:06:21,590 --> 01:06:23,950
But I don't have
to erase any edge.

1292
01:06:23,950 --> 01:06:25,770
Like, if I had to
erase this one,

1293
01:06:25,770 --> 01:06:27,950
that would be hard
because some motion here

1294
01:06:27,950 --> 01:06:31,080
might penetrate where that
edge ought to have been.

1295
01:06:31,080 --> 01:06:37,090
And erasing the
edge will make it--

1296
01:06:37,090 --> 01:06:39,320
adding the edge makes
it harder to fold.

1297
01:06:39,320 --> 01:06:41,880
So I can't erase it.

1298
01:06:41,880 --> 01:06:43,630
But the edges I have
to erase are the ones

1299
01:06:43,630 --> 01:06:45,610
that have been fully
inserted into the cone.

1300
01:06:45,610 --> 01:06:52,070
So if I can somehow do something
inside the cone, I would be OK.

1301
01:06:52,070 --> 01:06:55,610
All this work, defining a
canonical configuration,

1302
01:06:55,610 --> 01:06:58,586
was about forcing a chain to
stay inside a cone-- and not

1303
01:06:58,586 --> 01:07:00,210
only an alpha cone,
which is what we're

1304
01:07:00,210 --> 01:07:03,800
going to have as the ribosome,
but an alpha over two cone.

1305
01:07:03,800 --> 01:07:07,920
This is smaller than my
ribosome cone by factor of two.

1306
01:07:07,920 --> 01:07:10,452
And I need that.

1307
01:07:10,452 --> 01:07:11,820
Why do I need that?

1308
01:07:11,820 --> 01:07:17,957
Because this thing, I have no
control over the outside chain.

1309
01:07:17,957 --> 01:07:19,540
So the way that it
approaches the cone

1310
01:07:19,540 --> 01:07:22,800
could be as sharp
as like this, where

1311
01:07:22,800 --> 01:07:26,625
I have the first edge that
is-- the current edge that

1312
01:07:26,625 --> 01:07:27,516
is-- inside the cone.

1313
01:07:27,516 --> 01:07:28,890
As far as the
movie is concerned,

1314
01:07:28,890 --> 01:07:30,140
there's only one edge.

1315
01:07:30,140 --> 01:07:32,100
There's nothing up here.

1316
01:07:32,100 --> 01:07:35,020
I want to put something up
here, in a cone, naturally.

1317
01:07:38,030 --> 01:07:41,472
But I don't get to control
the first angle because that

1318
01:07:41,472 --> 01:07:42,680
is controlled by this motion.

1319
01:07:42,680 --> 01:07:45,247
Maybe it really needed
to go sharp like that

1320
01:07:45,247 --> 01:07:47,330
so that it could make a
sharper angle or whatever.

1321
01:07:52,720 --> 01:07:58,210
So you might say, well, of
course I can put it inside

1322
01:07:58,210 --> 01:08:03,020
and alpha cone, which is
the same as this alpha.

1323
01:08:03,020 --> 01:08:04,890
Sorry-- bad picture.

1324
01:08:04,890 --> 01:08:07,910
This is alpha.

1325
01:08:07,910 --> 01:08:10,970
This would be a problem because
I have this weird angle coming

1326
01:08:10,970 --> 01:08:14,480
in, and now suddenly I have
to bend back like that.

1327
01:08:14,480 --> 01:08:16,729
Maybe the turn angle here
is not so sharp as alpha.

1328
01:08:16,729 --> 01:08:19,620
Maybe it's one degree.

1329
01:08:19,620 --> 01:08:22,680
So I really can't force
the rest of my chain

1330
01:08:22,680 --> 01:08:25,210
to lie in this cone.

1331
01:08:25,210 --> 01:08:31,910
But I claim I can force
it to lie in this cone,

1332
01:08:31,910 --> 01:08:35,140
with half angle alpha over two.

1333
01:08:35,140 --> 01:08:38,800
This is a little more subtle.

1334
01:08:38,800 --> 01:08:39,990
Oh, right.

1335
01:08:39,990 --> 01:08:43,800
I wanted to mention that
helices appear in nature.

1336
01:08:43,800 --> 01:08:45,870
They appear in
proteins, but they also

1337
01:08:45,870 --> 01:08:49,063
appear in this
crazy climber plant.

1338
01:08:49,063 --> 01:08:50,729
Marty, have you seen
this in Costa Rica?

1339
01:08:50,729 --> 01:08:51,424
AUDIENCE: Yeah.

1340
01:08:51,424 --> 01:08:52,090
PROFESSOR: Yeah.

1341
01:08:52,090 --> 01:08:54,470
There's some really incredible
wildlife in Costa Rica.

1342
01:08:54,470 --> 01:08:56,678
I've never been there, but
I've seen lots of pictures

1343
01:08:56,678 --> 01:09:00,890
and this is spirals in practice.

1344
01:09:00,890 --> 01:09:03,645
Also, proteins tend
to form these things.

1345
01:09:03,645 --> 01:09:05,520
They call them out alpha
helices because they

1346
01:09:05,520 --> 01:09:08,979
spin like an alpha, I guess.

1347
01:09:08,979 --> 01:09:11,560
So it's kind of neat that the
canonical configuration, which

1348
01:09:11,560 --> 01:09:13,840
is totally
geometrically motivated,

1349
01:09:13,840 --> 01:09:16,591
also appears in biology.

1350
01:09:16,591 --> 01:09:19,710
Not that kind of biology though.

1351
01:09:19,710 --> 01:09:22,279
So here is the picture.

1352
01:09:22,279 --> 01:09:24,430
I have the big cone.

1353
01:09:24,430 --> 01:09:25,890
That is my ribosome,
and here I'm

1354
01:09:25,890 --> 01:09:30,170
going to write beta for
the-- beta for big, I guess.

1355
01:09:30,170 --> 01:09:34,240
And then we know that we
can canonicalize a chain,

1356
01:09:34,240 --> 01:09:36,560
or there at least exists
a canonical configuration

1357
01:09:36,560 --> 01:09:39,569
of the chain, where
everything lies

1358
01:09:39,569 --> 01:09:43,830
inside a cone of half
angle alpha over two.

1359
01:09:43,830 --> 01:09:48,560
Now the problem is that cone, we
want it to be at a funny angle.

1360
01:09:51,399 --> 01:09:52,790
Let me draw a real picture.

1361
01:09:59,780 --> 01:10:02,380
Here's a ribosome.

1362
01:10:02,380 --> 01:10:05,110
It has a big angle here.

1363
01:10:05,110 --> 01:10:08,120
We'll call alpha.

1364
01:10:08,120 --> 01:10:11,020
Now, I'm going to do
the not extreme case,

1365
01:10:11,020 --> 01:10:13,350
try to be a little more general.

1366
01:10:13,350 --> 01:10:15,490
There's some edge that
is currently entering,

1367
01:10:15,490 --> 01:10:20,040
and we have no control over that
edge or the rest of the chain.

1368
01:10:20,040 --> 01:10:21,770
That's determined
by the movie which

1369
01:10:21,770 --> 01:10:24,520
we're trying to play backwards.

1370
01:10:24,520 --> 01:10:28,136
What we have to control, and
what we're free to control,

1371
01:10:28,136 --> 01:10:29,760
is the rest of the
chain because as far

1372
01:10:29,760 --> 01:10:32,051
as the movie is concerned,
that hasn't been created yet

1373
01:10:32,051 --> 01:10:34,560
or it's already been destroyed,
depending on whether you're

1374
01:10:34,560 --> 01:10:36,814
playing forwards or backwards.

1375
01:10:36,814 --> 01:10:38,230
We need to say
what happens to it.

1376
01:10:38,230 --> 01:10:41,180
And what I want to happen
is so that by the time

1377
01:10:41,180 --> 01:10:45,100
this edge is inside,
that edge plus the rest

1378
01:10:45,100 --> 01:10:47,250
is in the canonical
configuration.

1379
01:10:47,250 --> 01:10:49,030
If I can achieve
that, then as edges

1380
01:10:49,030 --> 01:10:50,685
come in they become
canonicalized,

1381
01:10:50,685 --> 01:10:53,060
and then everything will be
canonical and inside the cone

1382
01:10:53,060 --> 01:10:54,490
and we're done.

1383
01:10:54,490 --> 01:10:56,370
That's our goal.

1384
01:10:56,370 --> 01:10:57,120
Canonicalization.

1385
01:10:57,120 --> 01:11:00,290
So, what's the deal?

1386
01:11:00,290 --> 01:11:07,240
Well, in reality,
there's some cone-- yeah,

1387
01:11:07,240 --> 01:11:09,430
it can penetrate
like that-- could

1388
01:11:09,430 --> 01:11:11,550
penetrate the outside cone.

1389
01:11:11,550 --> 01:11:13,230
This is getting messy.

1390
01:11:15,790 --> 01:11:20,950
So if I extend this line,
there's a cone of half angle

1391
01:11:20,950 --> 01:11:25,540
here, which is equal to
whatever the turn angle is

1392
01:11:25,540 --> 01:11:28,780
at that vertex, which
is, again, specified.

1393
01:11:28,780 --> 01:11:31,180
We're not free to set
it to whatever we want.

1394
01:11:31,180 --> 01:11:33,810
We know the next edge
must lie on this cone.

1395
01:11:36,460 --> 01:11:44,540
What I do-- now, on the
other hand, off to the side,

1396
01:11:44,540 --> 01:11:46,940
I have in mind a
vertical cone whose

1397
01:11:46,940 --> 01:11:48,900
half angle is alpha over two.

1398
01:11:48,900 --> 01:11:51,310
So it's quite small.

1399
01:11:51,310 --> 01:11:55,290
And I know-- how do
we do it-- initially,

1400
01:11:55,290 --> 01:11:58,970
the first edge was along
the maximum x direction,

1401
01:11:58,970 --> 01:12:00,670
and then it spirals
up from there.

1402
01:12:03,151 --> 01:12:03,650
OK.

1403
01:12:06,290 --> 01:12:07,880
Here's what I'm going to do.

1404
01:12:07,880 --> 01:12:09,940
There are sort of
two situations.

1405
01:12:09,940 --> 01:12:17,660
What I'd like to do is put a
cone here that is vertical.

1406
01:12:17,660 --> 01:12:20,010
Something like that.

1407
01:12:20,010 --> 01:12:21,590
Just like I have here.

1408
01:12:21,590 --> 01:12:24,780
The trouble is, the
right side of this cone

1409
01:12:24,780 --> 01:12:26,852
does not intersect the
boundary of this cone.

1410
01:12:26,852 --> 01:12:29,060
And need it to in order to
form the right angle here.

1411
01:12:29,060 --> 01:12:31,476
It's going to-- it might go
through the middle of the cone

1412
01:12:31,476 --> 01:12:32,357
like it does here.

1413
01:12:32,357 --> 01:12:34,690
So then what I do is I take
this canonical configuration

1414
01:12:34,690 --> 01:12:41,240
and I rotate it so that--
this is hard to draw-- it'll

1415
01:12:41,240 --> 01:12:42,445
be something like this.

1416
01:12:45,180 --> 01:12:47,690
There's no hope of seeing this.

1417
01:12:47,690 --> 01:12:49,420
So it's on the
surface of this cone.

1418
01:12:49,420 --> 01:12:50,720
It's not on the far right edge.

1419
01:12:50,720 --> 01:12:53,420
It's going to be some
intermediate point.

1420
01:12:53,420 --> 01:12:55,690
And that's exactly where
it intersects this cone.

1421
01:12:55,690 --> 01:12:58,890
It's just like the previous
picture, just harder to see.

1422
01:12:58,890 --> 01:13:01,120
I'm taking the intersection
of these two cones.

1423
01:13:01,120 --> 01:13:03,090
If I just rotate
the picture, then it

1424
01:13:03,090 --> 01:13:06,480
will lie in the intersection--
if they intersect.

1425
01:13:06,480 --> 01:13:07,700
But they might not intersect.

1426
01:13:10,750 --> 01:13:14,900
So let me go here
and try it again.

1427
01:13:19,270 --> 01:13:22,720
So the easy case is when
I can draw a vertical cone

1428
01:13:22,720 --> 01:13:27,480
and it intersects the cone
that I need to intersect.

1429
01:13:27,480 --> 01:13:33,950
The harder case is--
maybe it's more extreme.

1430
01:13:33,950 --> 01:13:36,130
Maybe it's a very
tight angle here.

1431
01:13:38,690 --> 01:13:41,390
So there's a very
small turn angle.

1432
01:13:41,390 --> 01:13:42,740
I have to intersect this cone.

1433
01:13:42,740 --> 01:13:44,590
Because the cone
that I'm working with

1434
01:13:44,590 --> 01:13:50,415
is actually smaller--
it's half of this angle--

1435
01:13:50,415 --> 01:13:51,790
it might not
intersect this cone.

1436
01:13:51,790 --> 01:13:54,950
In that case, I'm going
to rotate the cone

1437
01:13:54,950 --> 01:13:56,490
to fall over a little bit.

1438
01:13:59,740 --> 01:14:02,265
So instead of being like that,
it's going to be like this.

1439
01:14:07,870 --> 01:14:08,370
OK.

1440
01:14:08,370 --> 01:14:10,780
So I want the intersection
of these two cones, which

1441
01:14:10,780 --> 01:14:15,280
I've conveniently
made this edge here.

1442
01:14:15,280 --> 01:14:17,040
So the first edge
will lie along here,

1443
01:14:17,040 --> 01:14:18,920
and then it's going to
spiral out from there.

1444
01:14:18,920 --> 01:14:20,753
So I still have the
canonical configuration.

1445
01:14:20,753 --> 01:14:22,900
I've just tilted it.

1446
01:14:22,900 --> 01:14:25,720
Tilting is going to be necessary
because I have this angle

1447
01:14:25,720 --> 01:14:27,000
to match up.

1448
01:14:27,000 --> 01:14:31,270
The convenient thing about
the canonical-- the alpha

1449
01:14:31,270 --> 01:14:34,460
CCC, canonical
configuration-- is

1450
01:14:34,460 --> 01:14:38,940
that I have this half
angle of alpha over two,

1451
01:14:38,940 --> 01:14:43,800
so I can afford to tilt it
by up to alpha over two.

1452
01:14:43,800 --> 01:14:46,040
It will still stay
within the ribosome,

1453
01:14:46,040 --> 01:14:49,570
which is of half angle alpha.

1454
01:14:49,570 --> 01:14:54,340
And all I need to show here--
and once you think about it

1455
01:14:54,340 --> 01:14:57,770
for a while, it's
obvious-- that ideally, I

1456
01:14:57,770 --> 01:14:58,860
don't tilt it at all.

1457
01:14:58,860 --> 01:14:59,840
I've got tons of room.

1458
01:14:59,840 --> 01:15:01,600
Huge amount of room here.

1459
01:15:01,600 --> 01:15:04,330
But sometimes I'll have to
tilt it, but by at most alpha

1460
01:15:04,330 --> 01:15:04,830
over two.

1461
01:15:04,830 --> 01:15:07,300
So I will stay
inside the ribosome,

1462
01:15:07,300 --> 01:15:11,210
because this apex was inside
the cone and the smaller cone,

1463
01:15:11,210 --> 01:15:15,882
even if I tilt it all the
way to meet this edge,

1464
01:15:15,882 --> 01:15:17,090
it will stay inside the cone.

1465
01:15:17,090 --> 01:15:19,430
In fact, it'll stay
inside the big cone.

1466
01:15:19,430 --> 01:15:22,220
In fact, this cone that
I tilt-- the one that

1467
01:15:22,220 --> 01:15:24,560
contains the rest of the
canonical configuration--

1468
01:15:24,560 --> 01:15:29,840
will always contain
the up direction.

1469
01:15:29,840 --> 01:15:31,790
That's how to see it.

1470
01:15:31,790 --> 01:15:34,950
If it always contains the
up direction, then at most,

1471
01:15:34,950 --> 01:15:37,160
it's that big.

1472
01:15:37,160 --> 01:15:42,380
And that will-- because that's
an angle of alpha because that

1473
01:15:42,380 --> 01:15:44,370
was the half angle
of the big cone.

1474
01:15:44,370 --> 01:15:46,850
So that will be a half
angle of alpha over two.

1475
01:15:46,850 --> 01:15:49,630
As long as you contain the
up direction at all times,

1476
01:15:49,630 --> 01:15:53,140
you will not fall
outside the big cone.

1477
01:15:53,140 --> 01:15:56,810
So the rest is just momento.

1478
01:15:56,810 --> 01:15:59,420
So you play this
movie backwards.

1479
01:15:59,420 --> 01:16:02,690
As things come in
here, this cone

1480
01:16:02,690 --> 01:16:04,250
is going to wiggle
back and forth,

1481
01:16:04,250 --> 01:16:06,990
depending on how
this angle changes.

1482
01:16:06,990 --> 01:16:09,200
Once the edge gets
all the way in,

1483
01:16:09,200 --> 01:16:11,506
you absorb it into the
canonical configuration.

1484
01:16:11,506 --> 01:16:13,630
Little bit of work there
but you just sort of twist

1485
01:16:13,630 --> 01:16:16,749
the cone around
until that algorithm

1486
01:16:16,749 --> 01:16:19,040
that we described for producing
canonical configuration

1487
01:16:19,040 --> 01:16:21,370
would actually produce
what's inside the cone.

1488
01:16:21,370 --> 01:16:23,952
Then the next edge comes
in, cone wiggles around

1489
01:16:23,952 --> 01:16:25,410
until the edge gets
all the way in,

1490
01:16:25,410 --> 01:16:29,370
then you canonicalize what's
inside the cone, and repeat.

1491
01:16:29,370 --> 01:16:31,430
So if you had a
way to get it out,

1492
01:16:31,430 --> 01:16:32,960
you could put it
back in and keep

1493
01:16:32,960 --> 01:16:35,900
track of all the stuff
that happens on the inside.

1494
01:16:35,900 --> 01:16:38,180
That's what this theorem says.

1495
01:16:38,180 --> 01:16:43,290
And then there's just
slightly more to say.

1496
01:16:43,290 --> 01:16:50,110
If you have a flat
state, we want

1497
01:16:50,110 --> 01:16:52,680
to prove these things
are flat state connected.

1498
01:16:52,680 --> 01:16:55,900
So take some flat
state-- I really

1499
01:16:55,900 --> 01:16:59,220
should have only obtuse angles.

1500
01:16:59,220 --> 01:17:01,340
I claim this flat
state can be produced

1501
01:17:01,340 --> 01:17:05,210
using a cone of the
appropriate angle.

1502
01:17:05,210 --> 01:17:07,370
If the sharpest turn
angle here is alpha,

1503
01:17:07,370 --> 01:17:09,970
then you need an alpha cone.

1504
01:17:09,970 --> 01:17:11,510
And the way to
think about that is

1505
01:17:11,510 --> 01:17:17,041
to think of the cone moving
instead of as the chain moving.

1506
01:17:17,041 --> 01:17:17,790
It's a lot easier.

1507
01:17:17,790 --> 01:17:20,560
So you want the chain
to lie around here.

1508
01:17:20,560 --> 01:17:27,500
So you start by moving the
cone, I guess like this.

1509
01:17:27,500 --> 01:17:29,790
So that it just barely
touches the plane,

1510
01:17:29,790 --> 01:17:33,197
and this edge spews out.

1511
01:17:33,197 --> 01:17:34,863
Is that the right way
to think about it?

1512
01:17:34,863 --> 01:17:35,650
I don't know.

1513
01:17:35,650 --> 01:17:37,020
Should be like this.

1514
01:17:37,020 --> 01:17:38,780
So you take the cone.

1515
01:17:38,780 --> 01:17:41,740
This is like putting
frosting on a cake.

1516
01:17:41,740 --> 01:17:43,480
So you move your cone like here.

1517
01:17:43,480 --> 01:17:47,920
You squirt out some-- this edge.

1518
01:17:47,920 --> 01:17:51,910
Then, you do it so
that when you're here

1519
01:17:51,910 --> 01:17:56,120
and the new edge is created,
it's still in the plane.

1520
01:17:56,120 --> 01:17:58,550
And then you just sort
of move around there.

1521
01:17:58,550 --> 01:18:01,854
I'm just going to leave
it as a sketch like that.

1522
01:18:01,854 --> 01:18:03,270
Once you know that
you can produce

1523
01:18:03,270 --> 01:18:06,890
any flat state, of course,
you can reorient yourself

1524
01:18:06,890 --> 01:18:10,860
by relativity so that the chain
is moving instead of the cone.

1525
01:18:10,860 --> 01:18:13,410
So you can produce this
thing with a ribosome.

1526
01:18:13,410 --> 01:18:16,210
Once you know all flat
states can be produced

1527
01:18:16,210 --> 01:18:18,720
and you know all producible
configurations can

1528
01:18:18,720 --> 01:18:21,390
be canonicalized, then you
know it's flat state connected

1529
01:18:21,390 --> 01:18:24,010
and you know all canonical
things are flattenable and vice

1530
01:18:24,010 --> 01:18:27,900
versa, by continuous motions
without self intersection.

1531
01:18:27,900 --> 01:18:29,810
And all of this is algorithmic.

1532
01:18:29,810 --> 01:18:33,550
It can tell you how to go
from one place to another.

1533
01:18:33,550 --> 01:18:36,260
And so this is a
candidate algorithm,

1534
01:18:36,260 --> 01:18:38,790
I would say, for how
nature folds proteins.

1535
01:18:38,790 --> 01:18:40,410
Just thinking about
the mechanics,

1536
01:18:40,410 --> 01:18:42,330
not worrying about
how it's implemented.

1537
01:18:42,330 --> 01:18:45,100
Maybe you take
this model and then

1538
01:18:45,100 --> 01:18:50,450
you try to make it
physical forces,

1539
01:18:50,450 --> 01:18:52,220
and you get a way
to fold proteins.

1540
01:18:52,220 --> 01:18:54,120
That, of course,
remains a mystery.

1541
01:18:54,120 --> 01:18:58,800
But next time we will talk about
some very simple models that

1542
01:18:58,800 --> 01:19:02,800
are motivated more closely via
biology of how proteins might

1543
01:19:02,800 --> 01:19:05,340
actually fold, and talk
about the complexities

1544
01:19:05,340 --> 01:19:07,570
that you get there.