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[SQUEAKING]

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[RUSTLING]

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[CLICKING]

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MICHAEL SIPSER: Hi, everybody.

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I hope you can hear me.

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Give me a thumbs up if you can.

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Good.

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I see them coming through.

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So we're going to get started.

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So getting back to
what we've been doing.

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We've been talking
about space complexity.

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Measures how much memory
the algorithm requires

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or various problems require.

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And we defined the space
complexity classes,

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space f of n and
non-deterministic space

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f of n, the polynomial space
and non-deterministic space

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classes, and gave some
examples and so on.

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And today we're going to pick
up where we left off last time.

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One of the examples,
which is going

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to be an important one for
us, concerns this latter DFA

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problem.

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So I'm going to go over that
again, give a little bit more

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emphasis to the space
analysis, which I got

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some questions about last time.

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And then we are going
to move on from there

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and prove Savitch's
theorem and then

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talk about a complete
problems for PSPACE

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and show that this problem TQBF,
which we introduced last time,

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is actually a PSPACE
complete problem.

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But all in due course.

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A little bit of review.

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So we defined what we
mean by a Turing machine

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to run in a certain
amount of space.

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That means it uses
at most f of n.

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If it's running in space f of
n, uses at most f of n cells,

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tape cells, on every
input of length n.

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And similarly, a
non-deterministic Turing

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machine does the same.

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But in addition to that, the
non-deterministic machine

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has to halt on every
branch of its computation.

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And each branch
of its computation

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has to use at most that
bounded amount of tape cells.

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So we're going to be talking
about non-deterministic space

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computation today as well.

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It's going to be relevant to us.

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So we defined the
classes, as I mentioned,

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and the polynomial
and the PSPACE

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and non-deterministic
PSPACE classes.

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And this is how we believe
they relate to one another.

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The classes coNP and NP as well.

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And of course, as I
mentioned last time,

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there are some very major
unsolved problems in this area.

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So everything could
conceivably collapse down

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to P, which would, of
course, be very surprising.

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But we don't know how
to prove otherwise.

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And the big theorem that
we're going to prove today

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is that polynomial space and
non-deterministic polynomial

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space actually do collapse
down to each other.

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And being the same class.

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So in contrast with the
situation that we believe

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to be the case for
time complexity

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where we believe converting
non-deterministic

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to deterministic gives an
exponential increase for space

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complexity, it only gives
a squaring increase,

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as we'll see.

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So any questions on any of this?

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We will just march into a little
review of this latter problem.

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So reviewing some
of the notation.

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And let me emphasize that.

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So the big theorem we're
going to be proving today

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is that PSPACE and
NPSPACE are equal.

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And also we're
going to be talking

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about PSPACE completeness.

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But both of those
involve proving theorems.

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In the first case,
Savitch's theorem

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that converting
non-deterministic to

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deterministic
spaces is squaring.

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And in the second case, proving
that TQBF is PSPACE complete.

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Both of those theorems can be
thought of as generalizations

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of this theorem here that
the latter DFA problem can

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be done in the deterministic
polynomial space or n squared

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space.

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So it really pays
to try to understand

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how the proof of
this theorem works.

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Because in a sense, this theorem
is a more concrete version

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of what we're going to be seeing
in those other two theorems

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in a somewhat more
abstract form.

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So I like understanding things
in a more concrete way first.

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So that's why this is a good
example to start out with.

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But really in the
end of the day,

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it's the same
proof just repeated

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for those three theorems.

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So this is really three
for the price of one.

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Three theorems, one proof here.

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So you're going to be seeing
the same proof repeated

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three times but in different
levels of abstraction.

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So let's review again.

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I know some of you got it,
but maybe some of you didn't.

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And let's just try to be
clear on the algorithm

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to solve the ladder DFA problem.

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So if you remember, first of
all, let me just jump on ahead.

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The ladder problem is--

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a ladder, first of
all, is a sequence

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of strings that change one
symbol at a time that perhaps

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connect, go from one
string to another.

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So you're going to
go from work to play,

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changing one symbol at a time.

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So we gave an example
of this or you

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can easily come up with
an example of doing this.

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But the computational
problem is can you do it?

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Can you get from this
string to that string

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and stay within a
certain language?

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So it might be the
language of English words

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or it might be the
language of all strings

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that some specific
DFA recognizes.

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So it might be
all of the string.

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These might all be
strings that some DFA

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accepts or might be English
words or some other rule.

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And so that's what
we mean by trying

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to test if there's a ladder.

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And so the ladder problem
is, well, I don't think I

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wrote down the ladder
problem itself.

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But the bounded ladder problem
is basically the same idea.

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You're given the DFA.

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You're given the
strings u and v.

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And now this is
the bounded version

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of the problem where
I'm going to give you

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a limit on the number
of steps you can take.

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So I'm illustrating that here.

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So you're going to be given a b.

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And you want to say, can
I get from this string

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to that string within b steps?

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And we had a notation
for writing that.

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Going from u to v, if there's
a ladder that connects

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u to v within at most b steps.

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And so the bounded
ladder problem,

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which I'm introducing
because I'm

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going to be aiming toward
a recursive algorithm

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to solve this problem is can
I get from u to v by a ladder,

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changing one symbol at a time,
where each string along the way

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is accepted by b, and
I'm only allowed b steps.

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Little b steps.

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So that is the
computational problem

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that I'm going to be
solving with the algorithm

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that I'm going to describe.

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So the algorithm
I'm going to call

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BL for bounded ladder problem.

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And here is the input.

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And the algorithm
is, first of all,

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going to look to
see if b equals 1.

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If I'm just trying to get
from u to v in a single step.

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In that case, it's a
very simple problem,

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because you want to test,
obviously, that u and v are

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accepted by the automaton.

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And they just have to
differ in one place.

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And then you have a
very simple one step

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ladder that takes u to v.
So for the case b equals 1,

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it's very simple.

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For larger values
of b, we're going

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to solve the problem
recursively in terms

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of smaller values of b.

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So for b greater than 1, we're
going to recursively test.

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If you're trying to solve the
problem can I get from u to v,

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instead we're going to try
each possible halfway through.

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We don't know that
it's halfway through.

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So we're just going to
try each possible string.

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And we're going to test can we
get from u, the initial string,

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to that new string, that w,
in half the number of steps.

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And can I get to
the final string

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v in half the number of steps?

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If I can do that, then
I can get from u to v

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the total number of steps b.

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So I'm just going to
try to do this one w

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at a time for every possible w.

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This is going to be very
expensive in terms of time,

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but we're not worried
about time right now.

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We're trying to cut down
on the amount of space

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that we're using.

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And this is going to be
a big savings in space.

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Let's not worry about the
division, b over 2 here.

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All of the divisions,
we're going

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to be seeing this several times
going forward in the lecture,

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we'll think of them rounding up.

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But I'm not going to
make the notation look

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cumbersome by writing
that every time.

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OK.

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So here we go.

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Here is some candidate w string,
which is halfway through.

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Recursively test.

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Can I get from the starting
string to that w and from w

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to that ending string?

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If I can, if I find
such a w, then I accept.

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And if I try all possible w and
I never manage to find a way

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to make both the top
and the bottom work,

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then I know I cannot get
from the starting string

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to the ending string
within b steps.

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And so I reject.

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And now I'm going to solve
the original unbounded ladder

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problem by simply
putting the biggest

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possible bound into the
bounded ladder problem.

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And that's this value t, which
gives the very trivial bound

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of the total number
of possible strings

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that I can write down within my
length m that I'm working with.

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So this if sigma is the
alphabet of these strings,

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it's just sigma to the m.

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That's all possible strings.

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Of course, that's going to be
a maximum size on the ladder.

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So now how much
space does this take?

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And I think this is where
people got a little bit lost

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in the lecture last time.

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So I'm going to try
to animate this.

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I don't know if that's
going to help or not.

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But in the end of
the day, you're

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just going to have to think
through how do you account

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for the cost of this recursion.

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But the main thing,
to start off,

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you have to make sure you
understand the algorithm.

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And if you get
from here to there,

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we're going to try
all possible midpoints

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and then solve the upper part
and the lower part recursively,

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reusing the space.

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That's the way we're
going to get a saving.

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By solving this problem,
reusing the space

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that we use to
solve this problem.

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So I'm going to try to show
this to you on actually

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how the space gets used
on the Turing machine.

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You can kind of think
of here's the input

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and then after that is going to
be the stack for the recursion.

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If you're not that familiar
with how to implement recursion,

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it doesn't really matter.

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But you can just think about
what the algorithm needs

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to keep track of.

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And so as it's trying
every possible w,

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so just in order like
an odometer, just

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trying every possible
string, eventually maybe it

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finds a string that's
in the language that

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hears an English word, one
of the first English words

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of length 4 that
you might run into.

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And so now it makes sense,
actually, to do the recursion.

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So that's all.

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Every time you're going
to have to have a register

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or a location on the tape where
you're going to be writing down

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those different w's.

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So let's say it's over here.

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And we're just
going to go through.

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I hope that's not too
small for you to see.

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That's really where that
action is happening.

248
00:13:55,300 --> 00:13:57,480
And finally, maybe you
got to the string w.

249
00:13:57,480 --> 00:13:59,740
Now you're going to try
to do the recursion.

250
00:13:59,740 --> 00:14:06,930
So here as you're doing the
recursion on the top half

251
00:14:06,930 --> 00:14:08,440
again, you're going
to be cutting--

252
00:14:08,440 --> 00:14:13,830
you're going to be finding a
new w for the intermediate point

253
00:14:13,830 --> 00:14:17,848
just solving this upper
problem where we're testing

254
00:14:17,848 --> 00:14:19,140
if I can get from work to able.

255
00:14:19,140 --> 00:14:21,870
Later I'm going to have to deal
with getting from able to play.

256
00:14:29,010 --> 00:14:30,120
Good.

257
00:14:30,120 --> 00:14:34,650
So here, again, we're
going to be fixing able,

258
00:14:34,650 --> 00:14:38,010
fixing the first w.

259
00:14:38,010 --> 00:14:40,530
We're going to try
every possible way

260
00:14:40,530 --> 00:14:46,240
of getting from the start
string to that middle string.

261
00:14:46,240 --> 00:14:48,670
So we're going to try
every possible thing here.

262
00:14:48,670 --> 00:14:52,920
Eventually maybe we find some
other string in the language.

263
00:14:52,920 --> 00:14:55,920
We get down to the string book.

264
00:14:55,920 --> 00:15:00,420
And that's all
going to get stored.

265
00:15:00,420 --> 00:15:03,020
You can't forget
the string able.

266
00:15:03,020 --> 00:15:04,770
But now we're going
to use some more space

267
00:15:04,770 --> 00:15:08,160
to store those candidates.

268
00:15:08,160 --> 00:15:13,230
So that's a second version
of w deeper in the recursion.

269
00:15:13,230 --> 00:15:16,880
So here we're going to be
triangle to possible strings

270
00:15:16,880 --> 00:15:17,570
here.

271
00:15:17,570 --> 00:15:21,860
Again, eventually we
get to some string book.

272
00:15:21,860 --> 00:15:29,030
And if that succeeds in getting
us from work to able via book,

273
00:15:29,030 --> 00:15:31,820
now we're going to jump
down to do the bottom half,

274
00:15:31,820 --> 00:15:34,490
to see if I can get
from able to play

275
00:15:34,490 --> 00:15:39,150
as a separate problem, which
gets solved in the same space.

276
00:15:39,150 --> 00:15:43,850
So now here we're going to try
all these possibilities getting

277
00:15:43,850 --> 00:15:44,750
from able to play.

278
00:15:44,750 --> 00:15:47,970
Maybe call is the right
intermediate string there.

279
00:15:47,970 --> 00:15:51,662
And so now we're going
to erase the book

280
00:15:51,662 --> 00:15:53,412
and now we're going
to solve the lower sub

281
00:15:53,412 --> 00:15:55,440
problem in the same location.

282
00:15:55,440 --> 00:15:56,892
I hope this is helpful.

283
00:15:56,892 --> 00:15:58,850
This was a lot of work
making these animations.

284
00:16:05,970 --> 00:16:09,060
So the point of all
this is every time

285
00:16:09,060 --> 00:16:11,460
we go down the level
of the recursion,

286
00:16:11,460 --> 00:16:15,210
there's another
register whose size

287
00:16:15,210 --> 00:16:20,870
is big enough to hold one
of the strings, is needed.

288
00:16:20,870 --> 00:16:30,460
And that register
gets reused times

289
00:16:30,460 --> 00:16:35,410
throughout as we're going
through this recursion.

290
00:16:35,410 --> 00:16:39,820
So anyway, I hope
that's helpful.

291
00:16:39,820 --> 00:16:40,420
Anyway.

292
00:16:40,420 --> 00:16:46,450
So each level of the recursion
adds another order in

293
00:16:46,450 --> 00:16:48,100
to record the w.

294
00:16:48,100 --> 00:16:51,470
And so you have to do--
how many levels do we get.

295
00:16:51,470 --> 00:16:52,960
Well, the depth of
the recursion is

296
00:16:52,960 --> 00:16:55,630
going to be how
many times we end up

297
00:16:55,630 --> 00:17:01,120
having to divide this picture
in half until we get down to 1.

298
00:17:01,120 --> 00:17:04,569
And so the height of
this when we start off

299
00:17:04,569 --> 00:17:09,565
is going to be basically
an exponential in m. m

300
00:17:09,565 --> 00:17:11,410
is roughly the size of n.

301
00:17:11,410 --> 00:17:15,175
So when you take the log of
that, you're going to get--

302
00:17:18,950 --> 00:17:21,950
you're going to pull
down the exponential.

303
00:17:21,950 --> 00:17:24,109
So it's going to be
order m or, which is,

304
00:17:24,109 --> 00:17:26,150
again, roughly the
same size as the input.

305
00:17:26,150 --> 00:17:29,416
m is like half the input
because the whole input is u

306
00:17:29,416 --> 00:17:32,492
and v. m is just the size of u.

307
00:17:32,492 --> 00:17:36,140
And so each level
requires order n,

308
00:17:36,140 --> 00:17:39,230
and the depth is going
to be order n deep.

309
00:17:39,230 --> 00:17:44,630
The log of the initial
height of this ladder.

310
00:17:44,630 --> 00:17:49,480
And so the total space used is
going to be order n squared.

311
00:17:49,480 --> 00:17:51,600
So why don't we
just take a minute?

312
00:17:51,600 --> 00:17:54,510
I'm happy to spend a little
time going through this,

313
00:17:54,510 --> 00:17:57,990
either the algorithm's
correctness,

314
00:17:57,990 --> 00:18:00,690
understanding recursion,
or understanding the space

315
00:18:00,690 --> 00:18:01,260
analysis.

316
00:18:01,260 --> 00:18:02,802
If there's any
question that you feel

317
00:18:02,802 --> 00:18:07,950
you can ask that would be
clarifying for you, jump in.

318
00:18:07,950 --> 00:18:10,740
I'll set aside a few minutes
just to answer questions here.

319
00:18:13,640 --> 00:18:15,100
So I've got a question here.

320
00:18:15,100 --> 00:18:19,030
In step five, why
do we reject if all

321
00:18:19,030 --> 00:18:21,565
fail instead of just one fails?

322
00:18:25,520 --> 00:18:29,870
Well, here, so remember
what we're trying to do.

323
00:18:29,870 --> 00:18:34,110
We're trying to say can I
get from u to v in b steps?

324
00:18:36,910 --> 00:18:39,340
The way I'm going
to be doing that is

325
00:18:39,340 --> 00:18:42,490
trying every possible
intermediate string w.

326
00:18:45,750 --> 00:18:52,130
If I find some w
which does not work,

327
00:18:52,130 --> 00:18:53,980
that doesn't mean
that there's not

328
00:18:53,980 --> 00:18:55,630
some other w which might work.

329
00:18:55,630 --> 00:18:59,230
All I need is one
w for which I can

330
00:18:59,230 --> 00:19:02,230
get from u to w in half
the number of steps and w

331
00:19:02,230 --> 00:19:04,070
to v in half the
number of steps.

332
00:19:04,070 --> 00:19:06,160
So I'm going to try
every possible w.

333
00:19:06,160 --> 00:19:09,580
If any one of them is
good, then I can accept.

334
00:19:09,580 --> 00:19:12,370
If any one of them
succeeds where

335
00:19:12,370 --> 00:19:14,620
I can get from u to
w in half the steps

336
00:19:14,620 --> 00:19:17,875
and to w to v in half
the steps, then I

337
00:19:17,875 --> 00:19:22,070
know I can get from u to v
in the full number of steps.

338
00:19:22,070 --> 00:19:24,820
So I only need to find one.

339
00:19:24,820 --> 00:19:27,370
If one particular
one doesn't work,

340
00:19:27,370 --> 00:19:28,750
I'll just go on to the next one.

341
00:19:37,460 --> 00:19:37,975
OK.

342
00:19:37,975 --> 00:19:38,975
This is a good question.

343
00:19:41,820 --> 00:19:44,570
Do we have to save
the word book?

344
00:19:44,570 --> 00:19:47,750
So once we succeed in
getting from work to able,

345
00:19:47,750 --> 00:19:51,050
let's say via book, do
we need to save that word

346
00:19:51,050 --> 00:19:51,920
book anywhere?

347
00:19:51,920 --> 00:19:52,970
No.

348
00:19:52,970 --> 00:19:56,060
All we need to remember
is that we've succeeded

349
00:19:56,060 --> 00:19:58,550
in getting from work to able.

350
00:19:58,550 --> 00:20:00,780
We don't need to
remember book anymore.

351
00:20:00,780 --> 00:20:04,640
We just remember
that we've succeeded.

352
00:20:04,640 --> 00:20:09,560
And that is by virtue of
where we are in the algorithm.

353
00:20:09,560 --> 00:20:12,350
If we have succeeded,
then we move

354
00:20:12,350 --> 00:20:18,350
on to the second recursion,
second call, recursive call.

355
00:20:18,350 --> 00:20:20,700
So we found some way to do it.

356
00:20:20,700 --> 00:20:24,290
So we found some
intermediate point

357
00:20:24,290 --> 00:20:26,030
which succeeds for this one.

358
00:20:26,030 --> 00:20:27,320
So we move on to that one.

359
00:20:27,320 --> 00:20:29,430
We don't have to keep
any of that work anymore.

360
00:20:29,430 --> 00:20:31,160
All you have to do
is remember, yes,

361
00:20:31,160 --> 00:20:33,840
I can get from work to able
in half the number of steps.

362
00:20:33,840 --> 00:20:37,820
Now all that's left is
to get from able to play

363
00:20:37,820 --> 00:20:39,860
in half the number of steps.

364
00:20:39,860 --> 00:20:43,110
It doesn't matter how I got
to able in the first place.

365
00:20:43,110 --> 00:20:45,177
So we don't have
to remember that.

366
00:20:45,177 --> 00:20:46,510
That was a good question though.

367
00:20:49,300 --> 00:20:52,750
So I think I understand this.

368
00:20:52,750 --> 00:21:00,430
Before we replace the
value for book with call

369
00:21:00,430 --> 00:21:04,390
with the work
involved to find call,

370
00:21:04,390 --> 00:21:06,910
yeah, we have to check that
we can get from work to book

371
00:21:06,910 --> 00:21:08,810
and book to able.

372
00:21:08,810 --> 00:21:14,140
So we keep onto book while
we're working on the upper half.

373
00:21:14,140 --> 00:21:17,110
And only when we've
finally succeeded

374
00:21:17,110 --> 00:21:20,440
in getting from work to
able, let's say via book,

375
00:21:20,440 --> 00:21:24,380
then you can throw a book away.

376
00:21:24,380 --> 00:21:27,020
But while you're working on
the upper half, you try book,

377
00:21:27,020 --> 00:21:31,770
you try different strings of
length four until one of them

378
00:21:31,770 --> 00:21:32,270
works.

379
00:21:39,230 --> 00:21:39,800
I'm not sure.

380
00:21:39,800 --> 00:21:42,560
Somebody is asking me about
breadth first search and depth

381
00:21:42,560 --> 00:21:43,070
for search.

382
00:21:43,070 --> 00:21:44,300
I'm not sure I see it.

383
00:21:49,595 --> 00:21:51,970
I'm not sure that's going to
be a helpful way of thinking

384
00:21:51,970 --> 00:21:52,550
about this.

385
00:21:52,550 --> 00:21:54,300
So I'm not going to
answer that right now.

386
00:21:54,300 --> 00:22:00,730
But you can ask that
offline later if you want.

387
00:22:00,730 --> 00:22:06,340
Why is the recursion depth
log t instead of log m?

388
00:22:06,340 --> 00:22:08,500
Well, how high is this thing?

389
00:22:08,500 --> 00:22:13,280
Initially it's t high.

390
00:22:13,280 --> 00:22:16,310
But every time
we're doing a level,

391
00:22:16,310 --> 00:22:19,160
we're calling the recursion,
we're cutting t in half.

392
00:22:22,120 --> 00:22:24,070
I'm solving this
in general for b,

393
00:22:24,070 --> 00:22:28,980
but we starting off with b equal
to t. t is the maximum size.

394
00:22:28,980 --> 00:22:32,200
So initially this
is going to be t,

395
00:22:32,200 --> 00:22:35,200
and then it's going to be
t over 2, then t over 4.

396
00:22:35,200 --> 00:22:39,160
So it's going to be log t
levels before we get down to 1.

397
00:22:46,720 --> 00:22:47,220
Yeah.

398
00:22:47,220 --> 00:22:48,690
So somebody is
asking, can we think

399
00:22:48,690 --> 00:22:49,830
of this as a memory stack?

400
00:22:49,830 --> 00:22:51,247
Yes, this is like--
that's the way

401
00:22:51,247 --> 00:22:53,010
your typical
implementation of recursion

402
00:22:53,010 --> 00:22:56,010
is kind of with a stack, where
you push when you make a call

403
00:22:56,010 --> 00:22:57,780
and you pop when you
return from the call.

404
00:23:02,320 --> 00:23:08,990
Is it possible that v can
appear during BL procedure on t?

405
00:23:08,990 --> 00:23:12,652
Is it possible
that v can appear?

406
00:23:12,652 --> 00:23:13,860
I'm not sure what that means.

407
00:23:13,860 --> 00:23:16,200
It can reappear.

408
00:23:16,200 --> 00:23:19,127
So I'm starting with
u to v. Is it possible

409
00:23:19,127 --> 00:23:21,210
that v might be one of
these intermediate strings?

410
00:23:21,210 --> 00:23:21,710
Yeah.

411
00:23:25,020 --> 00:23:27,930
You're going to try every
possible intermediate stream

412
00:23:27,930 --> 00:23:29,250
blindly.

413
00:23:29,250 --> 00:23:30,510
Including v is one of them.

414
00:23:33,330 --> 00:23:39,030
If you can reach v more
quickly, well, great.

415
00:23:39,030 --> 00:23:41,550
I guess I have not dealt
with the issue of what

416
00:23:41,550 --> 00:23:43,875
happens if you get to a--

417
00:23:50,950 --> 00:23:52,750
technically it's
going to work out

418
00:23:52,750 --> 00:23:56,080
because I'm allowing
the difference

419
00:23:56,080 --> 00:23:57,590
to be in at most one place.

420
00:23:57,590 --> 00:23:59,290
So even if you get
there early, you're

421
00:23:59,290 --> 00:24:02,020
allowed to not change
anything, and that still

422
00:24:02,020 --> 00:24:03,670
is a legal step in the ladder.

423
00:24:08,280 --> 00:24:08,780
Yeah.

424
00:24:08,780 --> 00:24:11,710
I don't see how to do this
from a bottom up perspective.

425
00:24:11,710 --> 00:24:14,540
Somebody's is asking is there
a bottom up version of this.

426
00:24:14,540 --> 00:24:18,290
I don't think so.

427
00:24:18,290 --> 00:24:20,500
No, I don't think so.

428
00:24:20,500 --> 00:24:21,000
All right.

429
00:24:21,000 --> 00:24:23,710
Why don't we move on?

430
00:24:23,710 --> 00:24:28,050
So now we're going to
see this proof again.

431
00:24:28,050 --> 00:24:34,120
But this time we're
going to be proving

432
00:24:34,120 --> 00:24:39,460
that you can convert
any NFA to a DFA

433
00:24:39,460 --> 00:24:41,590
with only a squaring increase.

434
00:24:41,590 --> 00:24:50,600
So really, well, let me
just put that up there.

435
00:24:50,600 --> 00:24:53,270
So this is going to be
Savitch's theorem, that

436
00:24:53,270 --> 00:24:57,470
among other things proves
that PSPACE equals NPSPACE.

437
00:24:57,470 --> 00:25:02,000
So it says that you can convert
a non-deterministic machine

438
00:25:02,000 --> 00:25:05,553
to a deterministic machine only
squaring the amount of space.

439
00:25:05,553 --> 00:25:07,470
So you're comfortable
with this notation here.

440
00:25:07,470 --> 00:25:10,910
Anything that you can do in f
of n space non-deterministically

441
00:25:10,910 --> 00:25:15,860
you can do in f squared of
n based deterministically.

442
00:25:15,860 --> 00:25:17,660
And we're going
to accomplish that

443
00:25:17,660 --> 00:25:24,050
by converting an NTM
to a deterministic TM

444
00:25:24,050 --> 00:25:25,970
but only squaring
the space used.

445
00:25:25,970 --> 00:25:28,880
So n is going to
convert it to an m.

446
00:25:28,880 --> 00:25:31,250
And now this proof
is going to look

447
00:25:31,250 --> 00:25:35,890
very similar to the proof
in the previous slide.

448
00:25:35,890 --> 00:25:36,940
It's the same proof.

449
00:25:42,130 --> 00:25:45,850
And the fact from the
previous slide about ladder

450
00:25:45,850 --> 00:25:49,090
really is implied
by this, because we

451
00:25:49,090 --> 00:25:52,420
had an easy algorithm to show
that the latter problem is

452
00:25:52,420 --> 00:25:56,980
solvable in
non-deterministic, in NPSPACE.

453
00:25:56,980 --> 00:26:01,570
So ladder problem was
easily shown to be in here.

454
00:26:01,570 --> 00:26:06,300
If you remember, you
just basically guess

455
00:26:06,300 --> 00:26:07,830
the steps of the ladder.

456
00:26:07,830 --> 00:26:10,560
So non-deterministically,
you can easily check,

457
00:26:10,560 --> 00:26:13,500
can I get from the
start to the end?

458
00:26:13,500 --> 00:26:18,210
But Savitch's theorem
tells us that anything

459
00:26:18,210 --> 00:26:20,670
you can do non-deterministically
in polynomial space you

460
00:26:20,670 --> 00:26:22,870
can do deterministically
in polynomial space.

461
00:26:22,870 --> 00:26:25,320
So what we showed in
the previous slide

462
00:26:25,320 --> 00:26:28,080
follows from this
slide, but this slide

463
00:26:28,080 --> 00:26:30,390
is really just a generalization
of the same proof.

464
00:26:32,970 --> 00:26:35,550
Maybe I've said it
too many times now.

465
00:26:35,550 --> 00:26:40,140
So we're going to
introduce a notation very

466
00:26:40,140 --> 00:26:42,125
similar to the notation
we had last time.

467
00:26:42,125 --> 00:26:43,500
But now we're
going to be talking

468
00:26:43,500 --> 00:26:47,130
about simulating this
non-deterministic machine

469
00:26:47,130 --> 00:26:49,350
with a deterministic machine.

470
00:26:49,350 --> 00:26:51,570
And we're going to
take two configurations

471
00:26:51,570 --> 00:26:56,180
of this non-deterministic
machine, ci and cj,

472
00:26:56,180 --> 00:27:03,120
and say can I get from ci
to cj in at most b steps?

473
00:27:03,120 --> 00:27:05,160
I'm going to have a notation.

474
00:27:05,160 --> 00:27:08,580
Very similar to the notation
for the ladder where

475
00:27:08,580 --> 00:27:11,400
I can get from this word
to that word in at most b

476
00:27:11,400 --> 00:27:12,750
steps by a ladder.

477
00:27:12,750 --> 00:27:15,335
Here can I get from
this word to that--

478
00:27:15,335 --> 00:27:16,710
can I get from
this configuration

479
00:27:16,710 --> 00:27:18,600
to that configuration
with at most

480
00:27:18,600 --> 00:27:23,900
b steps of the Turing
machine's operation?

481
00:27:23,900 --> 00:27:28,540
So these are two
configurations now of n.

482
00:27:28,540 --> 00:27:32,260
So can n go from
this configuration ci

483
00:27:32,260 --> 00:27:35,710
to that other
configuration cj but only

484
00:27:35,710 --> 00:27:37,780
taking b steps along the way?

485
00:27:37,780 --> 00:27:39,820
That's now the
computational problem

486
00:27:39,820 --> 00:27:41,950
that I'm going to solve
for you with an algorithm.

487
00:27:41,950 --> 00:27:44,890
And it's going to be a recursion
exactly like the previous one.

488
00:27:47,940 --> 00:27:51,920
So n gets its input
the two configurations

489
00:27:51,920 --> 00:27:56,060
ci and cj and the bound
b and want to check

490
00:27:56,060 --> 00:27:59,620
can I get from i to j within b?

491
00:27:59,620 --> 00:28:04,430
So now the picture is a little
different but very similar.

492
00:28:04,430 --> 00:28:08,800
So instead of a
ladder appearing here,

493
00:28:08,800 --> 00:28:10,390
it's really something
that's basically

494
00:28:10,390 --> 00:28:16,580
a tableau for the
machine n where

495
00:28:16,580 --> 00:28:20,750
I have an initial configuration
and an ending configuration.

496
00:28:20,750 --> 00:28:27,505
This would happen to
be the starting point

497
00:28:27,505 --> 00:28:28,880
for the whole
procedure if you're

498
00:28:28,880 --> 00:28:31,890
testing whether n accepts w.

499
00:28:31,890 --> 00:28:35,390
But we would be solving
this in general for any--

500
00:28:35,390 --> 00:28:38,780
so that case, so I have the
start configuration of n on w

501
00:28:38,780 --> 00:28:42,650
and the accepting configuration
or an accepting configuration.

502
00:28:42,650 --> 00:28:48,420
But in general, what
I will be solving

503
00:28:48,420 --> 00:28:50,850
is starting with
any configuration ci

504
00:28:50,850 --> 00:28:54,160
and going to any
configuration cj.

505
00:28:54,160 --> 00:28:58,150
So I want to test
can I get from ci

506
00:28:58,150 --> 00:29:01,730
to cj within at most b steps.

507
00:29:01,730 --> 00:29:08,280
So first of all, if b is 1, you
can just check that directly.

508
00:29:08,280 --> 00:29:12,420
And now, again, we're
operating deterministically

509
00:29:12,420 --> 00:29:15,430
to simulate the
non-deterministic machine.

510
00:29:15,430 --> 00:29:17,760
So this is the
deterministic machine m.

511
00:29:17,760 --> 00:29:21,637
So m can easily, if it's
given two configurations of n

512
00:29:21,637 --> 00:29:23,220
and says, can I get
from the first one

513
00:29:23,220 --> 00:29:24,940
to the second one in one step?

514
00:29:24,940 --> 00:29:26,280
Well, that's an easy check.

515
00:29:26,280 --> 00:29:28,620
You just lay out those
two configurations,

516
00:29:28,620 --> 00:29:31,170
look at the n's transition
function, and say

517
00:29:31,170 --> 00:29:32,790
is this a legal move for n?

518
00:29:37,140 --> 00:29:37,800
Yes or no?

519
00:29:37,800 --> 00:29:40,700
And you accept or
reject accordingly.

520
00:29:40,700 --> 00:29:44,910
Now, if b is larger
than 1, you're

521
00:29:44,910 --> 00:29:52,340
going to try all possible
intermediate configurations,

522
00:29:52,340 --> 00:29:53,350
calling them c mid.

523
00:29:53,350 --> 00:29:57,400
This was like the w from
the previous theorem.

524
00:29:57,400 --> 00:30:00,130
This is all possible strings c--

525
00:30:00,130 --> 00:30:02,840
all possible
configurations c mid.

526
00:30:02,840 --> 00:30:06,170
And a configuration is
just going to be a--

527
00:30:11,370 --> 00:30:12,780
so far so-- OK.

528
00:30:12,780 --> 00:30:15,180
This is all possible.

529
00:30:15,180 --> 00:30:19,650
Looked like my PowerPoint
crashed, but it seems OK.

530
00:30:19,650 --> 00:30:21,970
This is all possible
configurations,

531
00:30:21,970 --> 00:30:24,420
which is just a string
with a string of tape

532
00:30:24,420 --> 00:30:26,610
symbols with a state
symbol appearing somewhere.

533
00:30:26,610 --> 00:30:28,420
That's all it is.

534
00:30:28,420 --> 00:30:34,720
Going to try all possible
configurations as candidate

535
00:30:34,720 --> 00:30:35,950
middle configurations.

536
00:30:35,950 --> 00:30:39,820
And say can I get from the upper
one to this candidate middle

537
00:30:39,820 --> 00:30:42,950
one and from that middle one
to the lower one within half

538
00:30:42,950 --> 00:30:44,200
the number of steps each time.

539
00:30:47,420 --> 00:30:49,336
And solving that
problem recursively.

540
00:30:56,320 --> 00:30:58,510
So I got a question here
about the possibility

541
00:30:58,510 --> 00:31:00,220
of looping forever.

542
00:31:03,310 --> 00:31:06,640
First of all, if n is
going to be looping,

543
00:31:06,640 --> 00:31:11,950
I don't have to worry about
it, because I'm starting off,

544
00:31:11,950 --> 00:31:16,360
I only need to simulate
machines that are deciders.

545
00:31:16,360 --> 00:31:18,730
Because I'm trying to show
that any language in here

546
00:31:18,730 --> 00:31:22,660
has to be accepted,
has to be decided

547
00:31:22,660 --> 00:31:25,823
by some non-deterministic
machine.

548
00:31:25,823 --> 00:31:28,240
So I'm not going to worry about
machines that are looping.

549
00:31:28,240 --> 00:31:33,820
If they're looping, m may
misbehave in some way,

550
00:31:33,820 --> 00:31:36,230
but that's not going
to be a problem for me.

551
00:31:36,230 --> 00:31:38,620
So let's keep life simple.

552
00:31:38,620 --> 00:31:41,350
Think about the deciders only.

553
00:31:44,800 --> 00:31:46,910
So we're going to
recursively test here.

554
00:31:46,910 --> 00:31:50,650
So that means I'm going to
try every possible middle,

555
00:31:50,650 --> 00:31:53,980
see if I can get from the start
to the middle and the middle

556
00:31:53,980 --> 00:31:56,500
to the end.

557
00:31:56,500 --> 00:32:02,110
If both of them work after
I test them recursively,

558
00:32:02,110 --> 00:32:03,280
then I'm going to accept.

559
00:32:03,280 --> 00:32:05,140
If not, I'm going to continue.

560
00:32:05,140 --> 00:32:08,410
And I reject if I try them all
and none of them have worked.

561
00:32:08,410 --> 00:32:12,310
Then I know there's no way for
n to get from this configuration

562
00:32:12,310 --> 00:32:14,500
to that configuration
in b steps.

563
00:32:18,580 --> 00:32:21,880
And the overall picture, I
test whether n accepts w,

564
00:32:21,880 --> 00:32:25,450
as I mentioned, by
starting with ci

565
00:32:25,450 --> 00:32:31,190
is the start configuration and
cj is the accept configuration.

566
00:32:31,190 --> 00:32:33,010
And now how big is t?

567
00:32:33,010 --> 00:32:38,560
Because I need to calculate
a bound on how deep

568
00:32:38,560 --> 00:32:41,950
the recursions are going to be.

569
00:32:41,950 --> 00:32:46,480
So t here is going to
be the total number

570
00:32:46,480 --> 00:32:48,940
of possible configurations.

571
00:32:48,940 --> 00:32:51,790
If this is the whole thing, it
never repeats a configuration.

572
00:32:51,790 --> 00:32:55,160
So this is going to be a
bound on how many steps

573
00:32:55,160 --> 00:32:56,650
then can be taken.

574
00:32:56,650 --> 00:33:00,220
And that's simply we
calculated this before.

575
00:33:00,220 --> 00:33:04,780
It's the number of
states times the number

576
00:33:04,780 --> 00:33:08,830
of head positions times the
number of tape contents.

577
00:33:08,830 --> 00:33:12,160
And this is really going to
be the dominant consideration

578
00:33:12,160 --> 00:33:13,040
anyway.

579
00:33:13,040 --> 00:33:17,595
And so now each recursive
level, and maybe I

580
00:33:17,595 --> 00:33:19,470
should have emphasized
this at the beginning,

581
00:33:19,470 --> 00:33:20,970
how wide is this picture?

582
00:33:20,970 --> 00:33:23,535
It's big enough to
store a configuration.

583
00:33:23,535 --> 00:33:27,360
A configuration is
essentially a tape contents.

584
00:33:27,360 --> 00:33:30,890
So that's going
to be f of n wide.

585
00:33:30,890 --> 00:33:34,360
So each recursive level
stores one configuration.

586
00:33:34,360 --> 00:33:39,490
Now the w costs f of
n space to write down.

587
00:33:39,490 --> 00:33:41,500
And the number of
levels is going

588
00:33:41,500 --> 00:33:45,640
to be the log of the initial
height, which is this.

589
00:33:45,640 --> 00:33:50,270
So this is going to be
the dominating part of it.

590
00:33:50,270 --> 00:33:53,710
So the log of this is going
to be, again, order f of n.

591
00:33:53,710 --> 00:33:56,830
So each one takes f of
n space to write down.

592
00:33:56,830 --> 00:33:59,920
The depth of the recursion
is going to be order f of n.

593
00:33:59,920 --> 00:34:02,470
So the total is going to
be order f squared of n,

594
00:34:02,470 --> 00:34:05,370
and that's how much
space this uses.

595
00:34:05,370 --> 00:34:09,000
And that's the proof
of Savitch's theorem.

596
00:34:09,000 --> 00:34:13,170
So yeah, so this is
a good point there.

597
00:34:13,170 --> 00:34:15,210
Somebody asks can
there be multiple

598
00:34:15,210 --> 00:34:17,250
accepting configurations.

599
00:34:17,250 --> 00:34:19,800
I should have made the--

600
00:34:19,800 --> 00:34:22,739
I forgot to say this and
I was just realizing it

601
00:34:22,739 --> 00:34:24,570
as I was explaining it.

602
00:34:24,570 --> 00:34:26,639
One of the things
I should have--

603
00:34:26,639 --> 00:34:30,090
you can enforce that there
is just a single accepting

604
00:34:30,090 --> 00:34:31,500
configuration.

605
00:34:31,500 --> 00:34:33,810
This is kind of a
detail, so don't worry

606
00:34:33,810 --> 00:34:37,260
about it if you don't want to.

607
00:34:37,260 --> 00:34:40,199
But you can make sure that
there's a single accepting

608
00:34:40,199 --> 00:34:44,130
configuration by telling
the machine when it accepts,

609
00:34:44,130 --> 00:34:47,040
it should erase its
tape and move its head

610
00:34:47,040 --> 00:34:50,110
into the left most cell in
the accept configuration.

611
00:34:50,110 --> 00:34:52,830
So there's just going to be a
single accepting configuration

612
00:34:52,830 --> 00:34:57,618
to worry about instead of having
to try multiple possibilities,

613
00:34:57,618 --> 00:34:59,160
which you could do
in this algorithm,

614
00:34:59,160 --> 00:35:00,960
but it would just
be annoying to have

615
00:35:00,960 --> 00:35:02,160
to write that down that way.

616
00:35:02,160 --> 00:35:04,140
So we often assume
there's just going

617
00:35:04,140 --> 00:35:05,883
to be a single
accepting configuration

618
00:35:05,883 --> 00:35:06,675
for these machines.

619
00:35:14,310 --> 00:35:17,580
How do you know f of n?

620
00:35:17,580 --> 00:35:22,840
So that's actually a little
bit of a delicate issue.

621
00:35:22,840 --> 00:35:25,980
I mean, if you could
compute f of n,

622
00:35:25,980 --> 00:35:28,470
the bound, which is,
for example, if it's

623
00:35:28,470 --> 00:35:31,680
going to be a polynomial bound,
you can just compute f of n.

624
00:35:31,680 --> 00:35:35,160
It's very easy to compute
n squared or n cubed.

625
00:35:35,160 --> 00:35:42,210
And so you can just
compute that and then use

626
00:35:42,210 --> 00:35:45,750
that as the size
of the registers

627
00:35:45,750 --> 00:35:48,720
you're going to be writing down.

628
00:35:48,720 --> 00:35:51,330
If you want to prove this
in general for f of n,

629
00:35:51,330 --> 00:35:54,270
it's a little bit technical
to have to deal with it.

630
00:35:54,270 --> 00:35:56,910
And I'm going to have to refer
you to the book on that one.

631
00:35:56,910 --> 00:35:59,820
The book tells you how solve
this for a general f of n.

632
00:35:59,820 --> 00:36:04,950
You basically have to try
every possible value from one

633
00:36:04,950 --> 00:36:07,000
until it works.

634
00:36:07,000 --> 00:36:14,310
And I'm afraid that's
going to derail

635
00:36:14,310 --> 00:36:16,680
us trying to decipher that.

636
00:36:16,680 --> 00:36:19,260
So let's not worry
about that aspect of it.

637
00:36:19,260 --> 00:36:23,082
But you can handle
general f of n here.

638
00:36:23,082 --> 00:36:25,040
You don't need to put
any conditions on f of n.

639
00:36:28,850 --> 00:36:32,640
Can we go over this term here?

640
00:36:32,640 --> 00:36:35,180
So we've seen this
term once before when

641
00:36:35,180 --> 00:36:40,850
we talked about LBAs and
seeing that LBAs always--

642
00:36:40,850 --> 00:36:42,920
we can solve the ALBA problem.

643
00:36:42,920 --> 00:36:45,470
This is simply the number
of different configurations

644
00:36:45,470 --> 00:36:48,140
the machine can have.

645
00:36:48,140 --> 00:36:50,390
Because the
configuration is a state.

646
00:36:50,390 --> 00:36:53,270
It's a head position and
a contents of the tape.

647
00:36:53,270 --> 00:36:56,330
And this is the number
of each of those

648
00:36:56,330 --> 00:36:58,280
that you can possibly have.

649
00:36:58,280 --> 00:37:00,380
Number of states.

650
00:37:00,380 --> 00:37:05,350
A head position, the size of
the tape of f of n is f of n.

651
00:37:05,350 --> 00:37:07,850
So this is that many
different head positions.

652
00:37:07,850 --> 00:37:09,550
And this is the number--

653
00:37:09,550 --> 00:37:12,220
if d is the size of
the tape alphabet,

654
00:37:12,220 --> 00:37:15,340
this is the number of tape
contents that you can have.

655
00:37:21,450 --> 00:37:24,690
How is seeing if n accepts
w with this algorithm

656
00:37:24,690 --> 00:37:26,490
convert a
non-deterministic Turing

657
00:37:26,490 --> 00:37:28,650
machine to some
deterministic Turing machine?

658
00:37:28,650 --> 00:37:31,560
Well, n is the
non-deterministic Turing.

659
00:37:31,560 --> 00:37:35,550
So n is we're converting
non-deterministic Turing

660
00:37:35,550 --> 00:37:40,050
machine n to deterministic
Turing machine m.

661
00:37:40,050 --> 00:37:42,930
So m is a deterministic
simulator of n.

662
00:37:42,930 --> 00:37:45,260
That's what this whole m is.

663
00:37:45,260 --> 00:37:51,020
So if we can do this for any n,
then we've proved our theorem.

664
00:37:55,590 --> 00:37:58,018
Why don't we defer.

665
00:37:58,018 --> 00:38:00,060
I think I got through most
of the questions here.

666
00:38:00,060 --> 00:38:01,920
If there's other
things, we can save them

667
00:38:01,920 --> 00:38:06,180
for the coffee break,
which is coming soon.

668
00:38:06,180 --> 00:38:10,050
I think we have one
more slide before that.

669
00:38:10,050 --> 00:38:12,640
So I'm going to define
PSPACE completeness I think.

670
00:38:12,640 --> 00:38:13,140
Yeah.

671
00:38:13,140 --> 00:38:18,010
And then I think after
that we have the break.

672
00:38:18,010 --> 00:38:25,040
So PSPACE completeness is
defined and very much inspired

673
00:38:25,040 --> 00:38:28,680
similarly to NP completeness.

674
00:38:28,680 --> 00:38:33,980
So a problem is PSPACE
complete if it's in NPSPACE

675
00:38:33,980 --> 00:38:38,420
and every other member of
PSPACE is reducible to it

676
00:38:38,420 --> 00:38:40,510
in polynomial time.

677
00:38:40,510 --> 00:38:44,530
And we'll say a bit about
why we choose polynomial time

678
00:38:44,530 --> 00:38:46,970
reducibility here.

679
00:38:46,970 --> 00:38:50,850
So here's kind of a
picture of how PSPACE--

680
00:38:50,850 --> 00:38:59,940
how complete problems relate
to their complexity classes.

681
00:38:59,940 --> 00:39:03,990
So you kind of think of a
complete problem for a class.

682
00:39:03,990 --> 00:39:06,420
It's kind of the hardest
problem in that class

683
00:39:06,420 --> 00:39:08,640
because you can convert.

684
00:39:08,640 --> 00:39:11,430
You can reduce any other
problem in that class

685
00:39:11,430 --> 00:39:13,060
to the complete problem.

686
00:39:13,060 --> 00:39:15,540
So here are the NP
complete problems.

687
00:39:15,540 --> 00:39:17,260
Sort of the hardest for NP.

688
00:39:17,260 --> 00:39:19,260
You have the PSPACE
complete problems,

689
00:39:19,260 --> 00:39:23,190
kind of the hardest for PSPACE.

690
00:39:23,190 --> 00:39:26,760
If an NP complete
problem goes into P,

691
00:39:26,760 --> 00:39:30,600
that pulls down all of NP to P.
If any PSPACE complete problem

692
00:39:30,600 --> 00:39:33,180
goes into P, it pulls
down all the PSPACE into P

693
00:39:33,180 --> 00:39:35,970
by following the
chain of reductions,

694
00:39:35,970 --> 00:39:38,070
because any PSPACE
problem is reducible

695
00:39:38,070 --> 00:39:39,400
to the complete problem.

696
00:39:39,400 --> 00:39:42,900
Which in turn if it's in P,
then everything goes into P.

697
00:39:42,900 --> 00:39:47,600
So if you have a PSPACE
complete problem which is in P,

698
00:39:47,600 --> 00:39:51,620
then all the PSPACE goes into P.

699
00:39:51,620 --> 00:39:57,290
So why do we use polynomial time
reducibility instead of, say,

700
00:39:57,290 --> 00:40:03,530
polynomial space reducibility
when we define this notion?

701
00:40:03,530 --> 00:40:06,030
It's kind of a very
reasonable question.

702
00:40:06,030 --> 00:40:09,350
But if you think about it, using
polynomial space reducibility

703
00:40:09,350 --> 00:40:10,550
would be a terrible idea.

704
00:40:14,260 --> 00:40:17,920
And we've seen this
phenomenon happen before.

705
00:40:17,920 --> 00:40:21,430
Every two problems
in PSPACE are going

706
00:40:21,430 --> 00:40:23,303
to be PSPACE reducible
to one another.

707
00:40:23,303 --> 00:40:24,970
We haven't even defined
that notion yet,

708
00:40:24,970 --> 00:40:26,980
but you can imagine
what it would be.

709
00:40:26,980 --> 00:40:32,080
Because a PSPACE reduction
can solve the problem

710
00:40:32,080 --> 00:40:34,390
for a problem in PSPACE.

711
00:40:34,390 --> 00:40:38,050
And then it can direct its
answer anywhere that it likes.

712
00:40:38,050 --> 00:40:42,070
So in general, when we
think about reductions,

713
00:40:42,070 --> 00:40:46,330
the reduction should be not
capable of solving the problems

714
00:40:46,330 --> 00:40:47,230
in the class.

715
00:40:47,230 --> 00:40:51,940
Because if they could, then
every two problems in the class

716
00:40:51,940 --> 00:40:53,913
would be reducible
to one another.

717
00:40:53,913 --> 00:40:55,330
And then all
problems in the class

718
00:40:55,330 --> 00:40:57,890
would be complete, because
everything in the class

719
00:40:57,890 --> 00:41:00,800
would be reducible to any
one of the other problems.

720
00:41:00,800 --> 00:41:02,590
So it would not be an
interesting notion.

721
00:41:02,590 --> 00:41:06,760
What you want to have happen
is that the reductions

722
00:41:06,760 --> 00:41:10,375
should be weaker than
the power of the class.

723
00:41:17,690 --> 00:41:20,240
And if you look
at the reductions

724
00:41:20,240 --> 00:41:23,720
that we've defined so far,
they're actually very simple.

725
00:41:23,720 --> 00:41:25,880
The only thing is they
have to make sure that they

726
00:41:25,880 --> 00:41:29,780
can make the output big enough.

727
00:41:29,780 --> 00:41:31,430
But actually
constructing the output,

728
00:41:31,430 --> 00:41:33,500
they're very simple
transformations.

729
00:41:33,500 --> 00:41:36,590
In fact, even polynomial time
is more than you typically need.

730
00:41:36,590 --> 00:41:38,750
There's even much
more limited classes

731
00:41:38,750 --> 00:41:42,450
that are capable of doing
the reductions, as we'll see.

732
00:41:42,450 --> 00:41:44,450
So having powerful
reductions is really

733
00:41:44,450 --> 00:41:47,270
not in the spirit of what
reductions are all about.

734
00:41:47,270 --> 00:41:51,980
You want very, very
simple transformations

735
00:41:51,980 --> 00:41:53,210
to be the reductions.

736
00:41:53,210 --> 00:41:55,400
Anyway, I hope that's helpful.

737
00:41:55,400 --> 00:41:58,100
So what we're going to be
aiming for in the second part

738
00:41:58,100 --> 00:42:03,200
of the lecture is showing
that TQBF is PSPACE complete.

739
00:42:03,200 --> 00:42:05,630
And let me-- here
is the check in.

740
00:42:10,020 --> 00:42:12,560
So this is our first
check in, coming

741
00:42:12,560 --> 00:42:14,040
a little late in the lecture.

742
00:42:14,040 --> 00:42:17,690
Suppose we have proven
that, as we will,

743
00:42:17,690 --> 00:42:21,130
that TQBF is PSPACE complete.

744
00:42:21,130 --> 00:42:26,290
What can we conclude if TQBF is
actually not necessarily in P,

745
00:42:26,290 --> 00:42:28,852
only goes to NP?

746
00:42:28,852 --> 00:42:30,560
And this is relevant
to a question that's

747
00:42:30,560 --> 00:42:33,810
coming in from the chat,
but I'll answer that later.

748
00:42:33,810 --> 00:42:38,390
So suppose TQBF
ends up being an NP

749
00:42:38,390 --> 00:42:40,940
and not in P. What
can we conclude?

750
00:42:40,940 --> 00:42:46,400
Remember, if TQBF
is in P, then PSPACE

751
00:42:46,400 --> 00:42:51,870
equals P. Suppose it goes to NP.

752
00:42:51,870 --> 00:42:54,700
What happens then?

753
00:42:54,700 --> 00:42:57,370
There may be several
correct answers here.

754
00:43:00,050 --> 00:43:01,910
Check all that apply.

755
00:43:05,270 --> 00:43:11,930
All right, so we're near
the end of the poll.

756
00:43:11,930 --> 00:43:14,348
So let me give you
another 10 seconds

757
00:43:14,348 --> 00:43:15,890
and then we're going
to shut it down.

758
00:43:20,510 --> 00:43:23,420
OK, are we all in?

759
00:43:23,420 --> 00:43:24,200
Closing it down.

760
00:43:27,520 --> 00:43:28,490
Here are the results.

761
00:43:28,490 --> 00:43:33,910
So yes, so first of all, the
most reasonable solution,

762
00:43:33,910 --> 00:43:38,740
most reasonable answer is
b, which I think most of you

763
00:43:38,740 --> 00:43:39,700
have gotten.

764
00:43:39,700 --> 00:43:45,250
That if a PSPACE complete
problem goes down to NP,

765
00:43:45,250 --> 00:43:48,790
well, NP is capable of
simulating the polynomial time

766
00:43:48,790 --> 00:43:50,380
reduction.

767
00:43:50,380 --> 00:43:55,210
And so any other
problem in PSPACE

768
00:43:55,210 --> 00:43:56,710
would then also be in NP.

769
00:43:56,710 --> 00:43:58,540
And PSPACE would equal NP.

770
00:43:58,540 --> 00:44:02,050
But note if PSPACE
equals NP, they're

771
00:44:02,050 --> 00:44:08,800
also NP equals coNP,
because PSPACE itself is

772
00:44:08,800 --> 00:44:11,300
closed under complementation.

773
00:44:11,300 --> 00:44:14,890
So that was kind of a
little bit extra fact

774
00:44:14,890 --> 00:44:17,960
that you could conclude
from this as well.

775
00:44:17,960 --> 00:44:22,420
So let's move on then
to our coffee break,

776
00:44:22,420 --> 00:44:28,180
and we'll pick up the proof
that TQBF is PSPACE complete

777
00:44:28,180 --> 00:44:29,200
after that.

778
00:44:43,860 --> 00:44:47,070
So was d true or not?

779
00:44:47,070 --> 00:44:48,660
D was P equal NP.

780
00:44:48,660 --> 00:44:57,930
No, we cannot conclude that
P equals NP from PSPACE equal

781
00:44:57,930 --> 00:45:00,480
to NP.

782
00:45:00,480 --> 00:45:05,190
So if TQBF is in NP, it
doesn't tell us anything.

783
00:45:05,190 --> 00:45:06,570
For all we know, P equals NP.

784
00:45:06,570 --> 00:45:11,710
But from the stuff
that we know so far,

785
00:45:11,710 --> 00:45:13,410
we cannot conclude
that P equals NP.

786
00:45:17,630 --> 00:45:21,350
Oh, and yeah, so you can
conclude-- oh, I'm sorry.

787
00:45:21,350 --> 00:45:25,190
B and d are both correct here.

788
00:45:27,800 --> 00:45:29,810
Let me just shut this thing off.

789
00:45:32,950 --> 00:45:34,330
So b and d are correct.

790
00:45:34,330 --> 00:45:37,750
So if a PSPACE complete
problem goes to NP,

791
00:45:37,750 --> 00:45:41,740
then NP equals
PSPACE, N equals coNP.

792
00:45:41,740 --> 00:45:44,810
So the correct answer, b and d.

793
00:45:44,810 --> 00:45:45,310
Sorry.

794
00:45:45,310 --> 00:45:47,380
I got myself confused.

795
00:45:47,380 --> 00:45:50,470
But c is not something
you can conclude or a.

796
00:45:55,640 --> 00:46:00,950
So somebody is asking
me a fair question.

797
00:46:00,950 --> 00:46:03,860
I say the reduction
method should be weaker

798
00:46:03,860 --> 00:46:07,400
than the class.

799
00:46:07,400 --> 00:46:10,430
But for example, even
in the case of PSPACE,

800
00:46:10,430 --> 00:46:14,030
PSPACE might be
equal to P. And then

801
00:46:14,030 --> 00:46:16,370
it wouldn't be
weaker than the class

802
00:46:16,370 --> 00:46:18,500
if we use polynomial
time reductions.

803
00:46:18,500 --> 00:46:22,880
But I think maybe I should
say apparently weaker.

804
00:46:22,880 --> 00:46:24,620
As far as we know, it's weaker.

805
00:46:24,620 --> 00:46:27,650
But we believe it to be weaker.

806
00:46:27,650 --> 00:46:32,060
It's true if P equals PSPACE,
then every problem in P

807
00:46:32,060 --> 00:46:34,010
is going to be PSPACE complete.

808
00:46:34,010 --> 00:46:36,740
It's just going to be a weird
world if P equals PSPACE.

809
00:46:36,740 --> 00:46:40,020
Same thing for NP and NP.

810
00:46:40,020 --> 00:46:42,060
So I'm getting a
number of questions

811
00:46:42,060 --> 00:46:45,330
also about other
possible reducibilities

812
00:46:45,330 --> 00:46:50,150
that are even weaker than
polynomial time reducibility.

813
00:46:50,150 --> 00:46:55,790
So we're going to see
very soon weaker reduce--

814
00:46:55,790 --> 00:46:59,630
complexity classes within
P. So first of all, PSPACE

815
00:46:59,630 --> 00:47:01,610
seems to be bigger
than P. We're also

816
00:47:01,610 --> 00:47:03,650
going to look at log space.

817
00:47:03,650 --> 00:47:07,100
But that's going to be
actually in Thursday's lecture.

818
00:47:07,100 --> 00:47:11,870
These are classes that
seem to be inside.

819
00:47:11,870 --> 00:47:14,120
Well, they're
inside P. We believe

820
00:47:14,120 --> 00:47:16,880
they're properly inside P.
But we'll see that later.

821
00:47:20,900 --> 00:47:22,270
Let me just see here.

822
00:47:25,290 --> 00:47:28,110
So we're almost out of time.

823
00:47:28,110 --> 00:47:30,580
Let me put our timer back.

824
00:47:30,580 --> 00:47:33,100
In fact, our timer is
showing us out of time.

825
00:47:33,100 --> 00:47:39,300
So why don't we get going?

826
00:47:39,300 --> 00:47:43,970
Let me move this back to--

827
00:47:46,830 --> 00:47:48,000
OK.

828
00:47:48,000 --> 00:47:48,780
Continuing.

829
00:47:51,300 --> 00:47:52,680
TQBF is PSPACE complete.

830
00:47:52,680 --> 00:47:54,240
So first of all,
let's remember TQBF.

831
00:47:58,330 --> 00:48:03,230
These are all of the quantified
Boolean formulas that are true.

832
00:48:03,230 --> 00:48:08,005
So TQBF stands for True
Quantified Boolean Formulas.

833
00:48:11,800 --> 00:48:13,940
And remember, we
saw these examples

834
00:48:13,940 --> 00:48:22,430
from the previous lecture that
these are two quantified--

835
00:48:22,430 --> 00:48:24,620
these are two QBFs.

836
00:48:24,620 --> 00:48:25,600
The first one is true.

837
00:48:25,600 --> 00:48:26,600
The second one is false.

838
00:48:26,600 --> 00:48:29,150
And it's going to be
interesting to think about.

839
00:48:29,150 --> 00:48:31,460
Here they're exactly
the same except

840
00:48:31,460 --> 00:48:33,840
for the order of
the quantifiers.

841
00:48:33,840 --> 00:48:37,770
And so what's really
going on here?

842
00:48:37,770 --> 00:48:41,550
I think it's good to
understand these expressions.

843
00:48:41,550 --> 00:48:47,610
They come up everywhere in
mathematics, these quantifiers.

844
00:48:47,610 --> 00:48:51,790
In the upper one, when
we say for every x,

845
00:48:51,790 --> 00:48:57,690
there exists a y, that y can
depend on the choice of x.

846
00:48:57,690 --> 00:48:59,640
You choose to make
a different x.

847
00:48:59,640 --> 00:49:02,220
You're allowed to
pick a different y.

848
00:49:02,220 --> 00:49:06,810
But the lower expression
says there's a universal y.

849
00:49:06,810 --> 00:49:12,080
There's one particular y
that works for every x.

850
00:49:12,080 --> 00:49:15,620
So in a sense, the
lower statement

851
00:49:15,620 --> 00:49:19,130
is a stronger statement.

852
00:49:19,130 --> 00:49:23,690
Whatever you have in the
quantifier free part.

853
00:49:23,690 --> 00:49:25,850
So the lower one
implies the upper one.

854
00:49:25,850 --> 00:49:28,310
Happens that the lower
one in this case is false

855
00:49:28,310 --> 00:49:29,540
and the upper one is true.

856
00:49:29,540 --> 00:49:34,070
But in general, when
you have this change

857
00:49:34,070 --> 00:49:38,510
of quantifiers like
this, the lower one

858
00:49:38,510 --> 00:49:40,850
would imply the upper one.

859
00:49:40,850 --> 00:49:42,530
Anyway, that's sort
of a side remark.

860
00:49:42,530 --> 00:49:47,000
So let's get back to the proof
that TQBF is PSPACE complete.

861
00:49:47,000 --> 00:49:50,600
That's what our goal is.

862
00:49:50,600 --> 00:49:52,100
All right.

863
00:49:52,100 --> 00:49:55,660
Now, as I mentioned,
this is the same proof.

864
00:49:55,660 --> 00:49:58,250
You're going to be seeing
it for the third time today.

865
00:49:58,250 --> 00:50:04,780
But there's a certain amount
of-- it's sort of the context

866
00:50:04,780 --> 00:50:08,210
changes in each case.

867
00:50:08,210 --> 00:50:11,930
So now we want to show that
TQBF is PSPACE complete.

868
00:50:11,930 --> 00:50:14,620
So it's one of these
hardest problems

869
00:50:14,620 --> 00:50:20,060
now but for PSPACE,
where satisfiability

870
00:50:20,060 --> 00:50:22,325
was the hardest problem for NP.

871
00:50:25,580 --> 00:50:30,590
So we want to show that
every language in PSPACE

872
00:50:30,590 --> 00:50:32,390
is reducible to TQBF.

873
00:50:35,660 --> 00:50:39,280
And so we're going to give
polynomial time reductions

874
00:50:39,280 --> 00:50:44,030
that map some particular problem
a, which can be done in space n

875
00:50:44,030 --> 00:50:44,530
to the k.

876
00:50:44,530 --> 00:50:48,440
It's a problem
solvable in PSPACE.

877
00:50:48,440 --> 00:50:53,120
We're going to show
how f maps a to TQBF.

878
00:50:53,120 --> 00:50:56,300
We have to construct the f.

879
00:50:56,300 --> 00:50:59,840
So f is going to be a
mapping that map strings

880
00:50:59,840 --> 00:51:01,580
which may or may not be in a.

881
00:51:01,580 --> 00:51:08,270
So strings w, which may or may
not be in a, to these formulas,

882
00:51:08,270 --> 00:51:11,430
these quantified formulas.

883
00:51:11,430 --> 00:51:17,540
So w is going to get mapped
to some formula phi sub mw.

884
00:51:17,540 --> 00:51:20,240
It had exactly the
same even symbols

885
00:51:20,240 --> 00:51:23,600
we used when the proof of
the Cook-Levin theorem about

886
00:51:23,600 --> 00:51:25,340
SAT being NP complete.

887
00:51:25,340 --> 00:51:29,300
This is a very similar proof.

888
00:51:29,300 --> 00:51:32,922
But you'll see that we have
to do something more in order

889
00:51:32,922 --> 00:51:34,130
to make it work in this case.

890
00:51:37,270 --> 00:51:43,890
So w is in a if and only if this
formula is going to be in TQBF.

891
00:51:43,890 --> 00:51:46,740
In other words, if and only
if this formula is true.

892
00:51:49,840 --> 00:51:54,810
So this formula is going
to kind of express the fact

893
00:51:54,810 --> 00:51:58,050
that m accepts w,
which means that w

894
00:51:58,050 --> 00:52:00,330
is in a, because
m is the machine,

895
00:52:00,330 --> 00:52:04,470
is the PSPACE machine for a.

896
00:52:04,470 --> 00:52:08,820
So this formula
says m accepts w,

897
00:52:08,820 --> 00:52:17,870
and it achieves that by building
in a simulation of m on w.

898
00:52:17,870 --> 00:52:21,260
It kind of describes a
simulation for m on w

899
00:52:21,260 --> 00:52:23,930
which ends up accepting.

900
00:52:23,930 --> 00:52:28,760
And if m does not accept
w, that description

901
00:52:28,760 --> 00:52:31,025
is going to inevitably be false.

902
00:52:34,710 --> 00:52:39,010
So let's just see what
that's going to look like.

903
00:52:39,010 --> 00:52:41,010
So we're going to
use the same idea

904
00:52:41,010 --> 00:52:44,730
that we used for the
Cook-Levin theorem

905
00:52:44,730 --> 00:52:46,440
that SAT is NP complete.

906
00:52:46,440 --> 00:52:49,050
This notion of a tableau.

907
00:52:49,050 --> 00:52:51,270
So if you remember,
it was basically

908
00:52:51,270 --> 00:52:56,830
a table which was just
simply a way of writing down

909
00:52:56,830 --> 00:53:01,290
a computation
history for m on w.

910
00:53:01,290 --> 00:53:06,060
So the rows are the
steps of the machine.

911
00:53:06,060 --> 00:53:08,640
The top row is the
start configuration.

912
00:53:08,640 --> 00:53:14,150
The bottom row is, let's
say, some particular accept

913
00:53:14,150 --> 00:53:16,970
configuration such as
I just described where

914
00:53:16,970 --> 00:53:21,350
the machine clears its tapes and
moves its head to the left end.

915
00:53:21,350 --> 00:53:23,272
So there's only one
accepting configuration

916
00:53:23,272 --> 00:53:24,230
we have to worry about.

917
00:53:27,700 --> 00:53:30,810
And each of the rows here
is a configuration of m.

918
00:53:37,090 --> 00:53:40,180
Because m runs in
space into the k,

919
00:53:40,180 --> 00:53:43,630
the tableau kind of similarly
to what I described before

920
00:53:43,630 --> 00:53:45,670
has width n to the k.

921
00:53:45,670 --> 00:53:50,570
So now we're talking about
polynomial time machines.

922
00:53:50,570 --> 00:53:53,920
So the f, which is the
bound, the space bound,

923
00:53:53,920 --> 00:53:56,500
is going to be some
polynomial n to the k.

924
00:53:56,500 --> 00:54:00,190
So the width of this
tableau, the size

925
00:54:00,190 --> 00:54:04,420
of these configurations
are going to be n to the k.

926
00:54:04,420 --> 00:54:07,700
How high is this
tableau going to be?

927
00:54:07,700 --> 00:54:11,050
Well, that's going to be
limited by the possible running

928
00:54:11,050 --> 00:54:19,330
time of the machine, which is
similar to what we saw before.

929
00:54:19,330 --> 00:54:23,050
It's going to be exponential
in the space bound.

930
00:54:23,050 --> 00:54:25,530
So it's going to be d
to the n to the k, where

931
00:54:25,530 --> 00:54:29,100
d is essentially the tape
alphabet of the machine.

932
00:54:32,230 --> 00:54:36,580
So are we all together on this?

933
00:54:36,580 --> 00:54:43,660
This is very similar to the
proof of SAT is NP complete.

934
00:54:43,660 --> 00:54:49,473
The key difference there
was m was non-deterministic,

935
00:54:49,473 --> 00:54:51,390
which might be something
to think about later.

936
00:54:51,390 --> 00:54:53,940
But let's not focus
on that right now.

937
00:54:53,940 --> 00:54:55,380
This m is deterministic.

938
00:54:58,650 --> 00:55:01,830
But the important difference
was the shape of the tableau.

939
00:55:01,830 --> 00:55:05,010
The size of the tableau
was very different.

940
00:55:05,010 --> 00:55:07,590
In the case of SAT
is NP complete,

941
00:55:07,590 --> 00:55:11,430
we started off with a polynomial
time non-deterministic machine.

942
00:55:11,430 --> 00:55:14,460
So it only could run for a
polynomial number of steps.

943
00:55:14,460 --> 00:55:16,320
Here is a polynomial
space machine,

944
00:55:16,320 --> 00:55:18,810
which can run for an
exponential number of steps.

945
00:55:18,810 --> 00:55:23,940
That's going to be an
important difference here.

946
00:55:23,940 --> 00:55:25,020
So let's see why.

947
00:55:29,320 --> 00:55:36,705
The reduction has to
construct this formula phi

948
00:55:36,705 --> 00:55:43,990
sub of mw, which basically
says that this tableau exists.

949
00:55:43,990 --> 00:55:46,180
Now, we already saw
how to do that when

950
00:55:46,180 --> 00:55:52,060
we proved the Cook-Levin
theorem that SAT is NP complete.

951
00:55:52,060 --> 00:55:53,620
Remember, we had all of those.

952
00:55:53,620 --> 00:55:57,100
We had variables for each
cell that told us what

953
00:55:57,100 --> 00:55:59,410
the contents of that cell was.

954
00:55:59,410 --> 00:56:01,580
And then we had a
lot of logic here.

955
00:56:01,580 --> 00:56:03,430
We had a bunch of
logic that said

956
00:56:03,430 --> 00:56:06,970
that all those
neighborhoods were correct,

957
00:56:06,970 --> 00:56:10,420
which basically says that
the tableau corresponds

958
00:56:10,420 --> 00:56:15,550
to a correct computation
of the machine.

959
00:56:15,550 --> 00:56:17,350
So why don't we just
do the same thing?

960
00:56:21,100 --> 00:56:24,550
Why don't we just
build our formula

961
00:56:24,550 --> 00:56:26,800
using exactly the
same process that we

962
00:56:26,800 --> 00:56:32,860
used to build the formula when
we had SAT being NP complete?

963
00:56:32,860 --> 00:56:34,450
Something goes wrong.

964
00:56:34,450 --> 00:56:37,610
We can't quite do that.

965
00:56:37,610 --> 00:56:41,470
The problem is that if
you remember the formula

966
00:56:41,470 --> 00:56:46,800
that we built before was really
about as big as the tableau is.

967
00:56:46,800 --> 00:56:49,830
Because it had some logic
for each one of the cells.

968
00:56:49,830 --> 00:56:52,140
It had a set of some
of the variables

969
00:56:52,140 --> 00:56:54,660
for each one of the cells
and it had some logic

970
00:56:54,660 --> 00:56:56,850
for each of those
neighborhoods, basically.

971
00:56:56,850 --> 00:57:00,660
So it says that each of the
cells does the right thing.

972
00:57:03,260 --> 00:57:07,300
So it was a pretty big formula,
but it was still polynomial.

973
00:57:07,300 --> 00:57:11,980
The problem is that tableau is
now n to the k by d to the n

974
00:57:11,980 --> 00:57:13,210
to the k.

975
00:57:13,210 --> 00:57:14,830
That's an exponentially
big object.

976
00:57:18,230 --> 00:57:21,940
So if your formula is going
to be as big as the tableau,

977
00:57:21,940 --> 00:57:24,850
there's no way you can hope
to produce that formula

978
00:57:24,850 --> 00:57:26,080
in polynomial time.

979
00:57:31,110 --> 00:57:32,660
And that's the problem.

980
00:57:32,660 --> 00:57:36,062
The formula is
going to be too big.

981
00:57:36,062 --> 00:57:38,020
Remember, we're trying
to get a polynomial time

982
00:57:38,020 --> 00:57:42,060
reduction from this language a.

983
00:57:42,060 --> 00:57:44,790
So we have an input
to a, a string that

984
00:57:44,790 --> 00:57:48,900
might be an a, which is
simulating the machine.

985
00:57:48,900 --> 00:57:53,670
And the size of the
tableau relative to w

986
00:57:53,670 --> 00:57:56,870
is going to be
something enormous.

987
00:57:56,870 --> 00:57:58,843
And so the formula is
as big as the tableau.

988
00:57:58,843 --> 00:58:01,260
There's no way to produce that
formula in polynomial time.

989
00:58:01,260 --> 00:58:05,390
So this is not going to work.

990
00:58:05,390 --> 00:58:06,770
Let's try again.

991
00:58:06,770 --> 00:58:07,990
So now we have here--

992
00:58:11,260 --> 00:58:16,045
now, so remember this notation
from ci to cj in b steps.

993
00:58:19,750 --> 00:58:25,470
So we're going to give a general
way of constructing formulas

994
00:58:25,470 --> 00:58:28,920
which express this
fact that I can

995
00:58:28,920 --> 00:58:35,280
get from configuration i of
m to configuration j of m

996
00:58:35,280 --> 00:58:37,310
in b steps.

997
00:58:37,310 --> 00:58:39,830
Whatever that b is. b is
going to be some bound.

998
00:58:39,830 --> 00:58:42,890
And I want to know can I
get from this configuration

999
00:58:42,890 --> 00:58:43,940
to that configuration.

1000
00:58:43,940 --> 00:58:46,700
And I want to write that
down as a formula, which

1001
00:58:46,700 --> 00:58:47,990
is going to express that fact.

1002
00:58:47,990 --> 00:58:49,407
And it'll be either
true or false.

1003
00:58:52,340 --> 00:58:56,170
And I'm going to give you
a recursive construction

1004
00:58:56,170 --> 00:58:57,080
for this formula.

1005
00:58:57,080 --> 00:59:01,810
So I'm going to build
that formula for a value

1006
00:59:01,810 --> 00:59:09,390
b out of formulas for
smaller values of b.

1007
00:59:09,390 --> 00:59:11,580
So this is going to be
a way of constructing

1008
00:59:11,580 --> 00:59:14,010
that formula in terms
of other formulas

1009
00:59:14,010 --> 00:59:15,150
that I'm going to build.

1010
00:59:15,150 --> 00:59:17,220
And there's going to be
a basis for the recursion

1011
00:59:17,220 --> 00:59:19,620
when b equals 1.

1012
00:59:19,620 --> 00:59:20,900
So that's the big picture.

1013
00:59:20,900 --> 00:59:23,050
So let's see how does
this formula look.

1014
00:59:27,110 --> 00:59:31,100
So let's not worry about the
case for b equal 1 right now.

1015
00:59:31,100 --> 00:59:35,500
This is the case for
larger values of b.

1016
00:59:35,500 --> 00:59:42,080
So the fact that I can get
from ci to cj within b steps.

1017
00:59:42,080 --> 00:59:45,600
I'm going to write
this down in this way.

1018
00:59:45,600 --> 00:59:52,400
And let's try to unpack that
and see what it's saying.

1019
00:59:56,050 --> 01:00:00,340
Without worrying about how are
we going to carry this out,

1020
01:00:00,340 --> 01:00:02,830
let's just try to understand
at a high level of semantics

1021
01:00:02,830 --> 01:00:05,890
of this thing what it's
trying to say to you.

1022
01:00:05,890 --> 01:00:10,880
It's going to say, well, I can
get from ci to cj in b steps.

1023
01:00:10,880 --> 01:00:14,750
So m can get from
ci to cj in b steps.

1024
01:00:14,750 --> 01:00:18,020
If there is some
other configuration c

1025
01:00:18,020 --> 01:00:20,860
mid, some other
configuration, I'm

1026
01:00:20,860 --> 01:00:27,100
calling it c mid, very much
inspired by the previous proof

1027
01:00:27,100 --> 01:00:28,790
of Savitch's
theorem, where there

1028
01:00:28,790 --> 01:00:31,430
was c mid was that
intermediate configuration.

1029
01:00:31,430 --> 01:00:33,730
So now instead of trying
them all, I'm saying

1030
01:00:33,730 --> 01:00:37,500
does there exist one
where I can get from ci

1031
01:00:37,500 --> 01:00:39,480
to c mid in half
the number of steps

1032
01:00:39,480 --> 01:00:42,120
and from c mid to cj in
half the number of steps?

1033
01:00:45,220 --> 01:00:50,100
So if I can build
these two formulas,

1034
01:00:50,100 --> 01:00:53,520
then I can combine them with
this sort of extra stuff

1035
01:00:53,520 --> 01:00:57,520
out here, and them together,
and put an exist quantifier

1036
01:00:57,520 --> 01:01:00,810
that says, does there exist
some configuration, some way

1037
01:01:00,810 --> 01:01:05,460
to find a configuration
such that it works

1038
01:01:05,460 --> 01:01:08,410
as these formulas require?

1039
01:01:08,410 --> 01:01:10,930
If I can do that, then
I'm going to be good.

1040
01:01:10,930 --> 01:01:13,810
Because then I can--

1041
01:01:13,810 --> 01:01:14,667
well, good.

1042
01:01:14,667 --> 01:01:16,750
At least I'll be good in
terms of making something

1043
01:01:16,750 --> 01:01:19,160
that is going to work.

1044
01:01:19,160 --> 01:01:21,280
So first of all,
let's understand

1045
01:01:21,280 --> 01:01:24,610
what I mean by writing down
does there exist c mid.

1046
01:01:24,610 --> 01:01:28,680
It's really if
you're thinking back

1047
01:01:28,680 --> 01:01:30,780
to the way we did the
Cook-Levin theorem,

1048
01:01:30,780 --> 01:01:33,750
we represented
these configurations

1049
01:01:33,750 --> 01:01:38,040
by variables which were
indicator variables for each

1050
01:01:38,040 --> 01:01:38,845
of the cells.

1051
01:01:38,845 --> 01:01:40,720
And we're going to do
exactly the same thing.

1052
01:01:40,720 --> 01:01:42,845
So we're going to have a
bunch of Boolean variables

1053
01:01:42,845 --> 01:01:47,810
which are going to represent
some configuration.

1054
01:01:47,810 --> 01:01:53,000
So more formally, or
in more detail, what

1055
01:01:53,000 --> 01:01:54,680
this does, there
exists a c mid, really

1056
01:01:54,680 --> 01:01:58,490
is an assignment to all
of those variables that

1057
01:01:58,490 --> 01:02:06,080
represent the contents of the
cells of that configuration.

1058
01:02:06,080 --> 01:02:10,670
OK, so now let's see how to--

1059
01:02:10,670 --> 01:02:12,440
how will the recursion work?

1060
01:02:12,440 --> 01:02:17,300
So to get this value
here, I'm going

1061
01:02:17,300 --> 01:02:19,640
to do the recursion further.

1062
01:02:19,640 --> 01:02:21,680
So does there exist a c mid?

1063
01:02:21,680 --> 01:02:26,870
And now for getting
from ci to this c mid,

1064
01:02:26,870 --> 01:02:28,310
is there some other c mid?

1065
01:02:28,310 --> 01:02:32,212
This is like another value of
w from the previous slide where

1066
01:02:32,212 --> 01:02:34,670
I'm getting, again, I'm cutting
the number of steps in half

1067
01:02:34,670 --> 01:02:35,170
again.

1068
01:02:35,170 --> 01:02:38,480
So going from b over 2
steps to b over 4 steps.

1069
01:02:38,480 --> 01:02:42,580
And I'm going to do the
same thing over here.

1070
01:02:42,580 --> 01:02:49,920
So I'm just unraveling
the construction

1071
01:02:49,920 --> 01:02:54,390
of this formula in
terms of building--

1072
01:02:54,390 --> 01:02:55,860
by building it up recursively.

1073
01:02:59,820 --> 01:03:03,450
And then I'm just going to keep
doing that until I get down

1074
01:03:03,450 --> 01:03:05,010
to the case where b equals 1.

1075
01:03:07,760 --> 01:03:12,410
And if I'm now trying to make
a formula that's going to be,

1076
01:03:12,410 --> 01:03:17,030
say I can get from ci to
cj in just a single step.

1077
01:03:17,030 --> 01:03:19,510
So this is really
talking about a tableau

1078
01:03:19,510 --> 01:03:22,180
of height 1 or height 2.

1079
01:03:22,180 --> 01:03:26,110
Then I can just directly
write that down the way

1080
01:03:26,110 --> 01:03:27,830
I-- now the tableau
is not very big.

1081
01:03:27,830 --> 01:03:30,430
So now I can write that
down using the neighborhoods

1082
01:03:30,430 --> 01:03:33,520
and so on that I did in the
Cook-Levin theorem proof.

1083
01:03:36,933 --> 01:03:38,600
And this is how you
put it all together.

1084
01:03:42,200 --> 01:03:49,400
And now if you want to
talk about the getting from

1085
01:03:49,400 --> 01:03:51,860
does m accept w.

1086
01:03:51,860 --> 01:03:54,530
So I initially say, can I get
from the start configuration

1087
01:03:54,530 --> 01:03:57,320
to the accept configuration
in t steps, which

1088
01:03:57,320 --> 01:03:59,130
is the maximum running
time of the machine?

1089
01:03:59,130 --> 01:04:00,110
So again.

1090
01:04:00,110 --> 01:04:03,920
And if you followed me, what
happened in Savitch's theorem,

1091
01:04:03,920 --> 01:04:06,140
it's the same values.

1092
01:04:08,860 --> 01:04:10,630
Now, the thing is we
have to understand

1093
01:04:10,630 --> 01:04:12,490
how big this formula is.

1094
01:04:12,490 --> 01:04:28,950
And if you think it
through, there's a problem.

1095
01:04:28,950 --> 01:04:30,770
Because what's going on here?

1096
01:04:30,770 --> 01:04:34,610
I'm expressing this formula
in terms of formulas

1097
01:04:34,610 --> 01:04:39,410
where the size of
b is cut in half.

1098
01:04:39,410 --> 01:04:41,710
But now there are two of them.

1099
01:04:41,710 --> 01:04:45,180
So it's two formulas
on half the value of b.

1100
01:04:45,180 --> 01:04:47,460
That's not going to
be a recursion that's

1101
01:04:47,460 --> 01:04:49,960
going to work in our favor.

1102
01:04:49,960 --> 01:04:52,030
So let's just see what happens.

1103
01:04:52,030 --> 01:04:59,160
So each recursive level
doubles the number of formulas.

1104
01:04:59,160 --> 01:05:00,990
Here we have two formulas.

1105
01:05:00,990 --> 01:05:05,340
Here we're going to have
four formulas and so on.

1106
01:05:05,340 --> 01:05:08,280
So the number of formulas
doubles each time.

1107
01:05:08,280 --> 01:05:12,080
So the length of the thing
that we're writing down

1108
01:05:12,080 --> 01:05:16,860
is doubling in size each time
we go down the recursion.

1109
01:05:16,860 --> 01:05:19,068
That's going to be OK if we
don't go too many levels.

1110
01:05:19,068 --> 01:05:21,193
But unfortunately, we are
going quite a few levels,

1111
01:05:21,193 --> 01:05:22,760
because the number
of levels is going

1112
01:05:22,760 --> 01:05:27,120
to be log of this initial
exponential size thing.

1113
01:05:27,120 --> 01:05:29,430
So it's going to be
n to the k levels.

1114
01:05:29,430 --> 01:05:31,970
And so if you're doubling
something n to the k times,

1115
01:05:31,970 --> 01:05:36,150
you're going to end up with an
exponentially sized formula.

1116
01:05:36,150 --> 01:05:40,430
And again, we failed.

1117
01:05:43,450 --> 01:05:45,880
OK, so I have a
check in on this.

1118
01:05:45,880 --> 01:05:47,470
But maybe we should
spend a couple

1119
01:05:47,470 --> 01:05:49,900
of minutes just
trying to understand

1120
01:05:49,900 --> 01:05:52,180
what's going on here.

1121
01:05:52,180 --> 01:05:55,030
Because the next slide
is really my last slide,

1122
01:05:55,030 --> 01:05:57,320
and it's going to
fix this problem.

1123
01:05:57,320 --> 01:06:00,700
But let's make sure we
understand what the problem is

1124
01:06:00,700 --> 01:06:02,260
before we try to fix it.

1125
01:06:07,260 --> 01:06:10,350
Why can we no longer write over
each layer of the recursion

1126
01:06:10,350 --> 01:06:11,280
as we did in ladder?

1127
01:06:11,280 --> 01:06:12,840
Oh, that's kind of
a cool question.

1128
01:06:12,840 --> 01:06:20,280
What does it even mean to write
over the different-- well,

1129
01:06:20,280 --> 01:06:23,500
so that's kind of an
interesting question here.

1130
01:06:23,500 --> 01:06:27,070
So in a sense, that's
going to be the solution.

1131
01:06:27,070 --> 01:06:35,020
We're going to reuse
things in a certain way.

1132
01:06:35,020 --> 01:06:42,010
But I want you to understand
that this method itself

1133
01:06:42,010 --> 01:06:45,760
does not work, because
this recursion here where

1134
01:06:45,760 --> 01:06:47,770
I'm writing the
formula, I'm building

1135
01:06:47,770 --> 01:06:51,130
the formula for
b out of formulas

1136
01:06:51,130 --> 01:06:55,000
for smaller values of b.

1137
01:06:55,000 --> 01:06:59,470
If I do it that way,
I'm going to end up--

1138
01:06:59,470 --> 01:07:03,310
if I do it as it's described
in this procedure here,

1139
01:07:03,310 --> 01:07:06,630
I'm going to end up with an
exponentially big formula.

1140
01:07:06,630 --> 01:07:10,110
And that's not good enough.

1141
01:07:10,110 --> 01:07:12,660
So if you cut the formula
in half each time,

1142
01:07:12,660 --> 01:07:14,460
even though you
have two formulas,

1143
01:07:14,460 --> 01:07:15,855
won't the length be the same?

1144
01:07:19,240 --> 01:07:20,740
I'm not cutting the
formula in half.

1145
01:07:20,740 --> 01:07:22,410
I'm cutting the
value of b in half.

1146
01:07:25,150 --> 01:07:26,530
So you have to
say b is initially

1147
01:07:26,530 --> 01:07:29,150
an exponentially big value.

1148
01:07:29,150 --> 01:07:32,810
So we're going to end up with
an exponentially big formula.

1149
01:07:32,810 --> 01:07:35,210
So it's not really cutting
the formula in half.

1150
01:07:35,210 --> 01:07:36,890
Cutting the b in half.

1151
01:07:36,890 --> 01:07:39,320
b starts out big.

1152
01:07:39,320 --> 01:07:43,320
I mean, b is initially
this value here,

1153
01:07:43,320 --> 01:07:44,460
d to the n to the k.

1154
01:07:47,480 --> 01:07:47,990
I'm worried.

1155
01:07:47,990 --> 01:07:49,157
Not too many questions here.

1156
01:07:49,157 --> 01:07:52,715
I have a feeling that's
probably not a good sign.

1157
01:07:57,980 --> 01:08:00,110
Well, I mean, if you're
hopelessly confused,

1158
01:08:00,110 --> 01:08:02,640
maybe I can't fix it quickly.

1159
01:08:02,640 --> 01:08:08,630
So anyway, why don't we move
on and see how to repair this,

1160
01:08:08,630 --> 01:08:09,740
how to fix this problem.

1161
01:08:12,950 --> 01:08:15,530
And that is going
to be by a trick.

1162
01:08:18,490 --> 01:08:21,147
In fact, I know the
people who were involved

1163
01:08:21,147 --> 01:08:22,189
with coming up with this.

1164
01:08:22,189 --> 01:08:27,370
This was actually, this proof
was done originally at MIT

1165
01:08:27,370 --> 01:08:30,160
many years ago in the 1970s.

1166
01:08:30,160 --> 01:08:32,710
And the folks who
were involved with it

1167
01:08:32,710 --> 01:08:34,840
called it the
abbreviation trick.

1168
01:08:34,840 --> 01:08:38,750
So that's what we're going
to do on the next slide.

1169
01:08:38,750 --> 01:08:41,800
Oh no, there's a check in first.

1170
01:08:41,800 --> 01:08:45,370
Why shouldn't we be surprised
that this construction fails?

1171
01:08:49,560 --> 01:08:51,450
A, well, we can't--

1172
01:08:51,450 --> 01:08:55,260
just the notion of defining
a quantified Boolean formula

1173
01:08:55,260 --> 01:09:01,710
by using recursion
is just not allowed.

1174
01:09:01,710 --> 01:09:06,590
So you can't define
formulas that way.

1175
01:09:06,590 --> 01:09:09,319
Doesn't use the for all
quantifier anywhere.

1176
01:09:09,319 --> 01:09:16,790
Or because we know that TQBF is
not in P. You can see we do--

1177
01:09:16,790 --> 01:09:18,470
what do you think?

1178
01:09:18,470 --> 01:09:24,510
Why should we not be surprised?

1179
01:09:24,510 --> 01:09:26,840
I guess I could have
put a d in there.

1180
01:09:26,840 --> 01:09:33,710
Not surprised because you
don't know what's going on.

1181
01:09:33,710 --> 01:09:35,540
That's another reason
not to be surprised.

1182
01:09:35,540 --> 01:09:40,700
But anyway, hopefully
you have some glimmer

1183
01:09:40,700 --> 01:09:42,979
of what's happening here.

1184
01:09:42,979 --> 01:09:45,200
And why don't we just--

1185
01:09:45,200 --> 01:09:47,069
almost finished.

1186
01:09:47,069 --> 01:09:51,220
So I'm going to shut
this down in a second.

1187
01:09:51,220 --> 01:09:51,774
Last call.

1188
01:09:54,410 --> 01:09:57,330
OK, ending.

1189
01:09:57,330 --> 01:09:59,650
All right.

1190
01:09:59,650 --> 01:10:03,700
So in fact, the
right answer is b.

1191
01:10:03,700 --> 01:10:09,280
I mean, one should be suspicious
that if there's no for alls

1192
01:10:09,280 --> 01:10:11,890
appearing in this
construction anywhere.

1193
01:10:11,890 --> 01:10:13,570
So really what
we're doing is we're

1194
01:10:13,570 --> 01:10:16,550
constructing a formula that
has only exist qualifiers.

1195
01:10:16,550 --> 01:10:18,650
So it's a
satisfiability problem.

1196
01:10:18,650 --> 01:10:22,810
So really what we just
did was we constructed--

1197
01:10:22,810 --> 01:10:28,360
we did in a more complicated
way the Cook-Levin construction,

1198
01:10:28,360 --> 01:10:31,750
because we end up
with a SAT formula

1199
01:10:31,750 --> 01:10:33,340
only with exist quantifiers.

1200
01:10:33,340 --> 01:10:37,577
And so really try one and
try two were the same.

1201
01:10:37,577 --> 01:10:39,160
So it's not surprising
that you end up

1202
01:10:39,160 --> 01:10:44,260
with an exponentially
big formula as a result.

1203
01:10:44,260 --> 01:10:45,430
I don't know.

1204
01:10:45,430 --> 01:10:47,920
A lot of you answered
we know that TQBF

1205
01:10:47,920 --> 01:10:51,100
is in P. It's not in
P. We don't know that.

1206
01:10:51,100 --> 01:10:53,800
I don't know where you--

1207
01:10:53,800 --> 01:10:57,280
what's happening with you guys?

1208
01:10:57,280 --> 01:10:58,540
But no.

1209
01:10:58,540 --> 01:11:00,490
Maybe that was a protest vote.

1210
01:11:00,490 --> 01:11:04,840
But anyway, we don't know that.

1211
01:11:04,840 --> 01:11:06,940
And what has that got to
do with anything anyway?

1212
01:11:06,940 --> 01:11:09,360
So anyway, let's see.

1213
01:11:09,360 --> 01:11:13,195
We solve this in our
remaining few minutes here.

1214
01:11:20,170 --> 01:11:21,265
So here is the solution.

1215
01:11:24,090 --> 01:11:28,365
Remember, this part where we're
saying we're trying to find c

1216
01:11:28,365 --> 01:11:31,802
mid, does there
exist a c mid such

1217
01:11:31,802 --> 01:11:34,260
that I can get from ci to c
mid in half the number of steps

1218
01:11:34,260 --> 01:11:37,260
and c mid to cj in half
the number of steps.

1219
01:11:37,260 --> 01:11:41,040
I'm expressing one formula
in terms of two formulas.

1220
01:11:41,040 --> 01:11:43,230
That's where the
blow up is occurring.

1221
01:11:43,230 --> 01:11:47,130
Because these two are then
in terms are going to each--

1222
01:11:47,130 --> 01:11:50,440
so these two are going to
become 4, become 8, and that's

1223
01:11:50,440 --> 01:11:50,940
not good.

1224
01:11:50,940 --> 01:11:54,750
Can I express this fact in
terms of just one formula?

1225
01:11:54,750 --> 01:11:57,870
And this kind of a little bit in
the spirit of your suggestions.

1226
01:11:57,870 --> 01:12:00,510
Can we kind of reuse
things in a way?

1227
01:12:00,510 --> 01:12:03,360
And that's what we're
actually going to kind do.

1228
01:12:03,360 --> 01:12:05,910
So here's another way
of saying the same thing

1229
01:12:05,910 --> 01:12:07,710
but with just a single formula.

1230
01:12:07,710 --> 01:12:09,270
And it uses a for all.

1231
01:12:09,270 --> 01:12:12,750
And the idea behind
it is that an and

1232
01:12:12,750 --> 01:12:14,850
is kind of like a for all.

1233
01:12:14,850 --> 01:12:17,070
Or a for all is
kind of like an and.

1234
01:12:17,070 --> 01:12:19,530
Just like in exist as
kind of like an or.

1235
01:12:19,530 --> 01:12:23,060
So when you're saying
does something exist,

1236
01:12:23,060 --> 01:12:25,760
is it this thing or that thing
or that thing or that thing?

1237
01:12:25,760 --> 01:12:27,530
And when you're
saying for all, it

1238
01:12:27,530 --> 01:12:31,570
has this thing and that thing
and that thing and that thing.

1239
01:12:31,570 --> 01:12:34,100
Ands and for alls are
very much related.

1240
01:12:34,100 --> 01:12:36,950
And so we're going to convert
this and into a for all.

1241
01:12:36,950 --> 01:12:42,820
We're going to say for all
configuration cg and ch that

1242
01:12:42,820 --> 01:12:50,380
are either set to ci c mid or
to c mid cj, the formula cg ch--

1243
01:12:50,380 --> 01:12:53,690
I can get from cg to ch in
half the number of steps.

1244
01:12:53,690 --> 01:12:55,930
So you have to think about
what this is meaning here.

1245
01:12:59,000 --> 01:13:00,910
And I also want to
make sure that you

1246
01:13:00,910 --> 01:13:04,630
don't feel I'm cheating you,
because well, first of all,

1247
01:13:04,630 --> 01:13:07,850
so now we have just
a single formula.

1248
01:13:07,850 --> 01:13:10,370
We're going to go down
to the case b equals 1,

1249
01:13:10,370 --> 01:13:12,300
as we did before.

1250
01:13:12,300 --> 01:13:17,490
You have to make sure that
saying this restricted for all

1251
01:13:17,490 --> 01:13:18,650
is not a cheat.

1252
01:13:18,650 --> 01:13:23,780
When you have for all x is
in s, like we have over here,

1253
01:13:23,780 --> 01:13:27,920
is equivalent to saying for
all x if x happens to be in s,

1254
01:13:27,920 --> 01:13:29,510
then the other thing follows.

1255
01:13:29,510 --> 01:13:34,960
And this implication can be
expressed using ands and ors.

1256
01:13:34,960 --> 01:13:39,910
And as before, the
initial starting point

1257
01:13:39,910 --> 01:13:44,870
is going to be going from c
start to c accept in t steps.

1258
01:13:44,870 --> 01:13:47,530
And so the analysis
that we get is

1259
01:13:47,530 --> 01:13:51,700
that because there's
no longer a blow up,

1260
01:13:51,700 --> 01:13:59,060
each recursive level just
adds this stuff here in front.

1261
01:13:59,060 --> 01:14:02,310
The exist c mid and
this for all part.

1262
01:14:02,310 --> 01:14:08,120
So that's going to be adding
order n to the formula instead

1263
01:14:08,120 --> 01:14:11,870
of multiplying because
we have two formulas.

1264
01:14:11,870 --> 01:14:14,810
And now the total number of
levels is order n to the k,

1265
01:14:14,810 --> 01:14:15,900
as before.

1266
01:14:15,900 --> 01:14:21,410
So the size is going to be n to
the k times n to the k order n

1267
01:14:21,410 --> 01:14:23,450
to the 2k.

1268
01:14:23,450 --> 01:14:26,490
I actually I had a
brief check in here,

1269
01:14:26,490 --> 01:14:31,850
which I'd like you to do just
in our remaining few seconds.

1270
01:14:31,850 --> 01:14:36,650
Does this construction depend
on m being deterministic?

1271
01:14:36,650 --> 01:14:39,380
So let me just launch that.

1272
01:14:39,380 --> 01:14:43,590
I want you to guys to get
your check in credit here.

1273
01:14:43,590 --> 01:14:53,880
But in fact, you have
to think this through.

1274
01:14:53,880 --> 01:14:59,485
That formula says that the
tableau, you get a tableau.

1275
01:14:59,485 --> 01:15:00,860
Is that going to
matter depending

1276
01:15:00,860 --> 01:15:03,350
upon whether it's deterministic
or non-deterministic?

1277
01:15:03,350 --> 01:15:09,840
It's actually, well, it's
kind of running 50/50 here.

1278
01:15:12,263 --> 01:15:13,680
Why don't you just
pick something?

1279
01:15:13,680 --> 01:15:17,220
Because I'd like to just
close this out and just

1280
01:15:17,220 --> 01:15:19,360
get to our last slide.

1281
01:15:19,360 --> 01:15:21,990
So if you don't
follow, don't worry.

1282
01:15:21,990 --> 01:15:27,270
But it's actually an interesting
point that the fact that this--

1283
01:15:27,270 --> 01:15:28,770
so I'm going to end this.

1284
01:15:28,770 --> 01:15:29,310
All in?

1285
01:15:33,750 --> 01:15:36,660
So in fact, the right
answer is it still would

1286
01:15:36,660 --> 01:15:38,790
work if it's non-deterministic.

1287
01:15:38,790 --> 01:15:40,920
And this would give
you an alternative way

1288
01:15:40,920 --> 01:15:42,510
of proving Savitch's theorem.

1289
01:15:42,510 --> 01:15:46,080
So really this all comes
down to this proof,

1290
01:15:46,080 --> 01:15:48,900
which implies Savitch's
theorem and then in turn

1291
01:15:48,900 --> 01:15:52,150
implies the ladder DFA problem.

1292
01:15:52,150 --> 01:15:55,320
So anyway, that's side note,
not critical for understanding,

1293
01:15:55,320 --> 01:15:55,830
really.

1294
01:15:55,830 --> 01:15:58,530
You can take those as all
separate, the results,

1295
01:15:58,530 --> 01:16:00,780
and that's good enough.

1296
01:16:00,780 --> 01:16:09,370
All right, so coming back.

1297
01:16:09,370 --> 01:16:10,450
Whoops.

1298
01:16:10,450 --> 01:16:11,020
Coming back.

1299
01:16:15,650 --> 01:16:17,480
This is what we did.

1300
01:16:17,480 --> 01:16:22,310
And so each recursive
level, the size of the QBF

1301
01:16:22,310 --> 01:16:23,265
is not the same.

1302
01:16:23,265 --> 01:16:25,640
Somebody's asking is it the
same at each recursive level.

1303
01:16:25,640 --> 01:16:27,200
No, we had to add in--

1304
01:16:33,530 --> 01:16:35,360
let's just see.

1305
01:16:35,360 --> 01:16:36,380
Each recursion.

1306
01:16:39,095 --> 01:16:41,990
This is recursively
built here, but now we

1307
01:16:41,990 --> 01:16:46,380
have to add this part in front
and this part here in front.

1308
01:16:46,380 --> 01:16:49,760
So the quantifier
which is quantified

1309
01:16:49,760 --> 01:16:51,710
over a bunch of
variables representing

1310
01:16:51,710 --> 01:16:55,320
the configuration, that
gets added on at each level.

1311
01:16:55,320 --> 01:16:58,373
So it's not just
it stays the same.

1312
01:16:58,373 --> 01:16:59,790
There's stuff
that's get added in.

1313
01:16:59,790 --> 01:17:05,470
But what's important is that it
doesn't blow up exponentially.

1314
01:17:05,470 --> 01:17:09,100
The stuff gets added in every
time but not multiplied.

1315
01:17:14,030 --> 01:17:14,530
OK.

1316
01:17:14,530 --> 01:17:15,770
So we're done here.

1317
01:17:15,770 --> 01:17:17,440
So anybody, you
can all take off.

1318
01:17:17,440 --> 01:17:19,690
I think many of
you already have.

1319
01:17:19,690 --> 01:17:21,920
Bye bye.

1320
01:17:21,920 --> 01:17:23,680
Thank you.