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PROFESSOR: All right, today we
are going to learn to count.

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One, two, three-- In a
algorithmic sense of course,

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and prove hardness of
counting style problems.

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Or more generally, any problem
where the output is an integer.

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Like approximation
algorithms, we

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need to define a slightly
stronger version of our NP

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style problem.

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It's not going to be
a decision problem.

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The remaining type
is a search problem.

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Search problem you're
trying to find a solution.

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This is just like
optimization problem,

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but with no objective function.

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Today all solutions
are created equal.

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So we just want to know
how many there are.

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So we're given an
instance, the problem.

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We want to produce a solution.

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Whereas in the
decision problem we

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want to know whether
a solution exists.

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With a search problem we
want to find a solution.

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Almost the same thing.

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00:01:56,040 --> 00:01:58,490
So that would be a search
problem in general.

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For an NP search problem you can
recognize what is an instance.

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And you could
recognize solutions

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to instances in polynomial time.

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Given an instance in the
solution you can say,

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yes that's a valid solution.

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OK for such a search
problem of course,

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is the corresponding
decision problem.

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Which is, does there
exist a solution?

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If you can solve this,
then you can solve that.

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You could solve the
decision problem in NP

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00:02:29,160 --> 00:02:32,460
by guessing a
solution and so on.

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00:02:32,460 --> 00:02:37,010
So this is intricately related
to the notion of a certificate

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for an NP problem.

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00:02:38,130 --> 00:02:40,250
The idea solutions
are certificates.

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00:02:40,250 --> 00:02:41,890
But when we say
problem is in NP,

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we say there is some way
to define certificate

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00:02:44,080 --> 00:02:47,730
so that this kind of
problem can be set up.

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00:02:47,730 --> 00:02:49,360
And the goal here--
the point here

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is to solidify a specific
notion of certificate.

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We can't just use any
one, because we're

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going to count them
the-- If you formulate

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00:02:55,550 --> 00:02:56,966
the certificates
in different ways

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00:02:56,966 --> 00:02:58,380
you'll get different counts.

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00:02:58,380 --> 00:03:03,350
But in general,
every NP problem can

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00:03:03,350 --> 00:03:09,210
be converted into an NP search
problem in at least one way.

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But each notional
certificate gives you

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00:03:15,880 --> 00:03:18,510
a notion of the search problem.

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OK, in some complexity context
these are called NP relations,

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00:03:22,700 --> 00:03:24,450
the way you specify
what a certificate is.

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00:03:24,450 --> 00:03:27,680
But I think this is the more
algorithmic perspective.

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All right, so
given such a search

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problem we turn it into
a counting problem.

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So that's a search
problem, is called A.

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Then counting problem will
be sharp A, not hashtag

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A. Some people call it
number A. The problem is,

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count the number of solutions
for a given instance.

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OK so in particular, you
detect whether the number

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is zero, or not zero.

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00:04:14,540 --> 00:04:16,850
So this is strictly
harder in some sense

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than the decision problem,
does there exist a solution.

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00:04:20,260 --> 00:04:23,670
And we will see some problems
where the search problem is

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polynomial, but the
corresponding counting problem

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is actually hard.

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I can't say NP
hard, there's going

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00:04:31,770 --> 00:04:34,310
to be a new notion of hardness.

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00:04:34,310 --> 00:04:36,760
So some examples.

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00:04:36,760 --> 00:04:40,630
Pretty much every problem we've
defined as a decision problem

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had a search problem in mind.

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So something like SAT, you need
to satisfy all of the things.

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00:04:47,607 --> 00:04:49,190
So there's no objective
function here,

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but you want to know
how many different ways,

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00:04:50,750 --> 00:04:53,200
how many different variable
assignments are there that

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00:04:53,200 --> 00:04:55,350
satisfy the given formula?

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00:04:55,350 --> 00:05:00,070
Or take your favorite
pencil and paper puzzle,

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00:05:00,070 --> 00:05:06,720
we'll be looking at
Shakashaka today, again.

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00:05:06,720 --> 00:05:08,540
How many different
solutions are there?

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00:05:08,540 --> 00:05:10,190
You'd like, when
designing a puzzle,

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00:05:10,190 --> 00:05:12,370
usually you want to
know that it's unique.

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00:05:12,370 --> 00:05:14,920
So it'd be nice if you could
count the number of solutions

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00:05:14,920 --> 00:05:16,930
and show that it's one.

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00:05:16,930 --> 00:05:20,700
These problems are going to turn
out to be very hard of course.

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00:05:20,700 --> 00:05:24,540
So let's define a
notion of hardness.

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00:05:24,540 --> 00:05:28,325
Sharp P is going to be the
class of all of these counting

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00:05:28,325 --> 00:05:28,825
problems.

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00:05:42,290 --> 00:05:44,095
This is the sort of certificate.

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00:05:44,095 --> 00:05:44,720
Yeah, question?

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00:05:44,720 --> 00:05:46,040
AUDIENCE: Just for the
puzzle application,

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00:05:46,040 --> 00:05:47,965
is it going to turn out
that counting if there's

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00:05:47,965 --> 00:05:50,048
one solution versus
one-to-one solution is as hard

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00:05:50,048 --> 00:05:51,898
as just counting total numbers?

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00:05:51,898 --> 00:05:56,350
PROFESSOR: It's
not quite as hard,

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00:05:56,350 --> 00:06:00,320
but we will show that
distinguishing one

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00:06:00,320 --> 00:06:04,430
for more than one is very
hard, is NP complete actually.

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00:06:04,430 --> 00:06:07,514
That's a decision problem that
could show that's NP complete.

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00:06:07,514 --> 00:06:09,180
So normally we think
of zero versus one,

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00:06:09,180 --> 00:06:11,060
but it turns out one
versus two is not--

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00:06:11,060 --> 00:06:15,870
or one versus more than one
is about the same difficulty.

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00:06:15,870 --> 00:06:19,290
Counting is even
harder I would say.

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00:06:19,290 --> 00:06:20,730
But it's bad news all around.

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00:06:20,730 --> 00:06:22,210
There's different
notions of bad.

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00:06:24,720 --> 00:06:29,810
Cool, so this is the sort
of certificate perspective.

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00:06:29,810 --> 00:06:32,190
With NP we-- I had given you
two different definitions

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00:06:32,190 --> 00:06:35,880
of certificate perspective, and
a non-deterministic computation

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00:06:35,880 --> 00:06:36,770
perspective.

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00:06:36,770 --> 00:06:40,290
You can do the same
computational perspective here.

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00:06:40,290 --> 00:06:47,290
You could say sharp P
is the set of problems,

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00:06:47,290 --> 00:06:51,165
solved by polynomial time.

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00:06:55,660 --> 00:06:57,995
Call it a non-deterministic
counting algorithm.

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00:07:03,297 --> 00:07:05,380
We don't need Turing
machines for this definition,

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00:07:05,380 --> 00:07:08,040
although that was of course
the original definition.

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00:07:08,040 --> 00:07:11,320
So take your favorite
non-deterministic algorithm

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00:07:11,320 --> 00:07:14,920
as usual for NP, it makes
guesses at various points,

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00:07:14,920 --> 00:07:16,580
multiple branches.

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00:07:16,580 --> 00:07:20,720
With the NP algorithm the way
we'd execute it on an NP style

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00:07:20,720 --> 00:07:22,970
machine is that we would
see whether there's

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00:07:22,970 --> 00:07:24,910
any path that led to a yes.

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00:07:24,910 --> 00:07:27,790
Again, it's going to output
yes or no at the end.

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00:07:27,790 --> 00:07:30,930
In this case, what the computer
does, the sharp P style

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00:07:30,930 --> 00:07:34,290
computer, is it-- conceptually
it runs all the branches.

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00:07:34,290 --> 00:07:37,310
It counts the number of yeses
and returns that number.

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00:07:37,310 --> 00:07:41,210
So even though the algorithm is
designed to return yes or no,

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00:07:41,210 --> 00:07:44,030
when it executes it
actually outputs a number.

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00:07:44,030 --> 00:07:47,190
The original paper
says magically.

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00:07:47,190 --> 00:07:50,014
It's just as magic as an NP
machine, but a little-- even

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00:07:50,014 --> 00:07:50,930
a little more magical.

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00:07:53,630 --> 00:07:57,080
OK so you-- I mean if you're not
comfortable with that we just

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00:07:57,080 --> 00:08:00,330
use this definition, same thing.

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00:08:00,330 --> 00:08:07,727
This is all work done by Les
Valiant in the late '70s.

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00:08:07,727 --> 00:08:08,310
These notions.

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00:08:22,710 --> 00:08:24,320
So it's pretty clear,
it's pretty easy

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00:08:24,320 --> 00:08:26,410
to show all these
problems are in sharp P,

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00:08:26,410 --> 00:08:28,951
because they were-- the
corresponding decision problems

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00:08:28,951 --> 00:08:29,450
were in NP.

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00:08:29,450 --> 00:08:34,110
We convert them into
sharp P algorithms.

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00:08:34,110 --> 00:08:37,210
Now let's think about hardness
with respect to sharp P.

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00:08:37,210 --> 00:08:38,990
As usual we want
it to mean as hard

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00:08:38,990 --> 00:08:40,409
as all problems in that class.

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00:08:40,409 --> 00:08:45,270
Meaning that we can reduce all
those problems to our problem.

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00:08:45,270 --> 00:08:48,520
And the question is, by
what kind of reductions?

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00:08:48,520 --> 00:08:51,785
And here we're going to allow
very powerful reductions.

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00:08:56,972 --> 00:08:58,430
We've talked about
these reductions

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00:08:58,430 --> 00:09:01,320
but never actually been
allowed to use them.

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00:09:01,320 --> 00:09:03,550
Multicall, Cook-style
reductions.

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00:09:06,080 --> 00:09:08,770
I think in general, this
is a common approach

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00:09:08,770 --> 00:09:14,530
for FNP, which is Functions
NP style functions.

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00:09:14,530 --> 00:09:17,520
Which you can also think of
this as kind of-- the counting

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00:09:17,520 --> 00:09:19,850
version is, the
output is a value.

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00:09:19,850 --> 00:09:24,050
So you have a function instead
of just a decision question.

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00:09:24,050 --> 00:09:28,190
When the output is some thing,
some number, in this case.

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00:09:28,190 --> 00:09:30,790
You might have to manipulate
that number at the end.

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00:09:30,790 --> 00:09:33,180
And so at the very least,
you need to make a call

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00:09:33,180 --> 00:09:35,360
and do some stuff before
you return your modified

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00:09:35,360 --> 00:09:37,290
answer in your reduction.

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00:09:37,290 --> 00:09:39,870
But in fact we're going
to allow full tilt,

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00:09:39,870 --> 00:09:43,790
you could make multiple calls
to some hypothetical solution

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00:09:43,790 --> 00:09:48,757
to your problem in order to
solve all problems in sharp P.

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00:09:48,757 --> 00:09:50,840
And we'll actually use
multicall a bunch of times.

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00:09:50,840 --> 00:09:53,140
We won't always need multicall.

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00:09:53,140 --> 00:09:55,850
Often we'll be able to get
away with a much simpler kind

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00:09:55,850 --> 00:09:56,910
of reduction.

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00:09:56,910 --> 00:09:58,570
Let me tell you that kind now.

170
00:10:01,990 --> 00:10:03,881
But in general we allow
arbitrary multicall.

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00:10:03,881 --> 00:10:04,380
Yeah?

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00:10:04,380 --> 00:10:07,820
AUDIENCE: Your not limited
in the number of multicalls?

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00:10:07,820 --> 00:10:08,550
PROFESSOR: Right.

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00:10:08,550 --> 00:10:10,580
You could do polynomial
number of multicalls.

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00:10:10,580 --> 00:10:14,030
As before, reduction
should be polynomial time.

176
00:10:14,030 --> 00:10:18,070
But, so you're basically
given an algorithm.

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00:10:18,070 --> 00:10:24,160
That's usually called an Oracle
that solves your problem,

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00:10:24,160 --> 00:10:28,960
solves your problem B. And
you want to solve-- yeah,

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00:10:28,960 --> 00:10:33,030
you want to solve A by
multiple calls to B. So

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00:10:33,030 --> 00:10:35,370
what, we'll see a bunch
of examples of that.

181
00:10:35,370 --> 00:10:39,370
Here's a more familiar
style of reduction.

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00:10:39,370 --> 00:10:42,060
And often, for a lot of problems
we can get away with this.

183
00:10:42,060 --> 00:10:43,950
But especially a lot
of the early proofs

184
00:10:43,950 --> 00:10:45,380
needed the multicall.

185
00:10:45,380 --> 00:10:47,380
And as you'll see you can
do lots of cool tricks

186
00:10:47,380 --> 00:10:50,500
with multicall
using number theory.

187
00:10:53,060 --> 00:10:56,790
So parsimonious reduction.

188
00:10:56,790 --> 00:10:58,855
This is for NP search problems.

189
00:11:03,470 --> 00:11:09,080
This is going to be a lot like
an NP reduction, regular style.

190
00:11:09,080 --> 00:11:16,470
So again we convert an instance
x of a, five function f,

191
00:11:16,470 --> 00:11:22,700
into an instance x prime of b.

192
00:11:26,880 --> 00:11:32,030
And that function should
be computable in poly time.

193
00:11:34,970 --> 00:11:37,386
So far just like
an NP reduction.

194
00:11:40,060 --> 00:11:43,450
And usually we would say,
for a search problem,

195
00:11:43,450 --> 00:11:45,630
we would say there
exists a solution

196
00:11:45,630 --> 00:11:48,740
for x, if and only if, there
exists a solution for x prime.

197
00:11:48,740 --> 00:11:53,670
That would be the direct analog
of an NP style reduction.

198
00:11:53,670 --> 00:11:55,670
But we're going to ask
for a stronger condition.

199
00:11:55,670 --> 00:12:01,040
Which is that the number
of solutions to problem

200
00:12:01,040 --> 00:12:10,890
A to instance x, equals the
number of solutions of type B

201
00:12:10,890 --> 00:12:13,540
to instance x prime.

202
00:12:13,540 --> 00:12:16,050
OK so in particular, this
one's going to equal to one,

203
00:12:16,050 --> 00:12:18,520
this one will be greater than
equal to one, and vice versa.

204
00:12:18,520 --> 00:12:20,829
So this is stronger
than an NP style

205
00:12:20,829 --> 00:12:22,870
reduction for the
corresponding decision problem.

206
00:12:27,820 --> 00:12:29,780
Yeah, so I even wrote that down.

207
00:12:29,780 --> 00:12:34,100
This implies that the decision
problems have the same answer.

208
00:12:42,100 --> 00:12:45,435
So in particular, this implies
that we have an NP reduction.

209
00:12:52,020 --> 00:12:59,540
So in particular if A,
the decision version of A

210
00:12:59,540 --> 00:13:02,680
is NP hard, then the decision
version of B is NP hard.

211
00:13:02,680 --> 00:13:10,400
But more interesting is
that, if A is sharp P hard,

212
00:13:10,400 --> 00:13:12,130
then B is sharp P hard.

213
00:13:17,820 --> 00:13:21,080
But also this holds for NP.

214
00:13:21,080 --> 00:13:25,130
For the decision versions
of A and B. Sorry,

215
00:13:25,130 --> 00:13:30,505
sharp A and sharp B. OK this
is a subtle distinction.

216
00:13:33,600 --> 00:13:36,710
For sharp P hardness we're
talking about the counting

217
00:13:36,710 --> 00:13:38,630
problems, and we're
talking about making calls

218
00:13:38,630 --> 00:13:40,067
to other counting solutions.

219
00:13:40,067 --> 00:13:42,400
Then doing things with those
numbers and who knows what,

220
00:13:42,400 --> 00:13:44,290
making many calls.

221
00:13:44,290 --> 00:13:45,930
With parsimonious
reduction we're

222
00:13:45,930 --> 00:13:48,990
thinking about the
non-counting version.

223
00:13:48,990 --> 00:13:50,830
Just the search problem.

224
00:13:50,830 --> 00:13:55,050
And so we're not worried about
counting solutions directly,

225
00:13:55,050 --> 00:13:57,300
I mean it-- what's nice
about parsimonious reductions

226
00:13:57,300 --> 00:13:59,040
is they look just
like NP reductions

227
00:13:59,040 --> 00:14:01,250
for the regular old problems.

228
00:14:01,250 --> 00:14:03,240
We just need this
extra property,

229
00:14:03,240 --> 00:14:07,180
parsimony that the number
of solutions to the unit

230
00:14:07,180 --> 00:14:09,210
is preserved through
the transformation.

231
00:14:09,210 --> 00:14:14,350
And a lot of the proofs
that we've covered follow--

232
00:14:14,350 --> 00:14:17,160
have this property and
will be good for us.

233
00:14:17,160 --> 00:14:19,410
If we can get our
source problems hard,

234
00:14:19,410 --> 00:14:22,430
then we'll get a lot of
target problems hard as well.

235
00:14:25,390 --> 00:14:34,100
Well let me tell you about one
more version which I made up.

236
00:14:34,100 --> 00:14:35,305
C-monious reductions.

237
00:14:43,550 --> 00:14:45,060
This is my attempt
at understanding

238
00:14:45,060 --> 00:14:47,760
the entomology of parsimonious.

239
00:14:47,760 --> 00:14:53,880
Which is something like little
money, being very thrifty.

240
00:14:53,880 --> 00:14:56,030
So this is having a
little bit more money.

241
00:14:56,030 --> 00:14:58,047
You have C dollars.

242
00:14:58,047 --> 00:14:59,880
But you have to be very
consistent about it.

243
00:14:59,880 --> 00:15:01,920
I should probably add
some word that means

244
00:15:01,920 --> 00:15:03,110
uniform in the middle there.

245
00:15:03,110 --> 00:15:08,390
But I want the number of
solutions of x prime to equal

246
00:15:08,390 --> 00:15:13,315
c times the number
of solutions to x.

247
00:15:16,110 --> 00:15:19,080
C a fixed constant.

248
00:15:19,080 --> 00:15:22,510
And it has to be the
same for every x.

249
00:15:25,700 --> 00:15:28,020
This would be just as good
from a sharp P perspective.

250
00:15:28,020 --> 00:15:31,690
Because if I could solve
B and I wanted to solve A,

251
00:15:31,690 --> 00:15:35,680
I would convert A to B, run
the thing, then divide by C.

252
00:15:35,680 --> 00:15:39,830
There will never be a remainder
and then I have my answer to A.

253
00:15:39,830 --> 00:15:41,690
As long as C's not zero.

254
00:15:41,690 --> 00:15:45,930
C should be an integer here.

255
00:15:45,930 --> 00:15:51,790
We will see a bunch of
C-monious reductions.

256
00:15:51,790 --> 00:15:58,110
I guess, yeah it doesn't have
to be totally independent of x.

257
00:15:58,110 --> 00:16:00,480
It can depend on things like n.

258
00:16:00,480 --> 00:16:03,870
Something that we can
compute easily I guess.

259
00:16:03,870 --> 00:16:06,030
Shouldn't be too dependent on x.

260
00:16:06,030 --> 00:16:10,340
All right, let's do-- let's
look at some examples.

261
00:16:10,340 --> 00:16:23,030
So, going to make a list here
of sharp P complete problems.

262
00:16:25,830 --> 00:16:29,560
And we'll start with versions
of SAT because we like SAT.

263
00:16:29,560 --> 00:16:36,120
So I'm just going to tell
you that sharp 3SAT is hard.

264
00:16:36,120 --> 00:16:40,490
First sharp SAT is hard, and
the usual proof of SAT hardness

265
00:16:40,490 --> 00:16:45,460
shows sharp P completeness
for sharp SAT.

266
00:16:45,460 --> 00:16:49,340
And if you're careful about
the conversion from SAT to 3CNF

267
00:16:49,340 --> 00:16:51,290
you can get sharp
three satisfied.

268
00:16:51,290 --> 00:16:53,240
It's not terribly
interesting and tedious.

269
00:16:53,240 --> 00:16:57,660
So I will skip that one.

270
00:16:57,660 --> 00:17:01,120
So what about Planar 3SAT?

271
00:17:03,710 --> 00:17:06,970
I stared at this diagram
many times for a while.

272
00:17:06,970 --> 00:17:12,130
This is lecture seven for
replacing a crossing in a 3SAT

273
00:17:12,130 --> 00:17:14,210
thing with this picture.

274
00:17:14,210 --> 00:17:17,950
And all of this argument and
this table in particular,

275
00:17:17,950 --> 00:17:20,599
was concluding
that the variables

276
00:17:20,599 --> 00:17:21,900
are forced in this scenario.

277
00:17:21,900 --> 00:17:25,089
If you know what
A and B are-- so

278
00:17:25,089 --> 00:17:26,650
once you choose
what A and B are,

279
00:17:26,650 --> 00:17:28,650
these two have to
be copies of A and B

280
00:17:28,650 --> 00:17:31,760
and then it ended up
that A-- We proved

281
00:17:31,760 --> 00:17:34,680
that A1 equaled A2
and B1 equaled B2,

282
00:17:34,680 --> 00:17:36,860
and then these
variables were all

283
00:17:36,860 --> 00:17:38,820
determined by these formula.

284
00:17:38,820 --> 00:17:41,440
And so once you know A and
B all the variable settings

285
00:17:41,440 --> 00:17:42,220
are forced.

286
00:17:42,220 --> 00:17:45,180
Which means you preserve
the number of solutions.

287
00:17:45,180 --> 00:17:50,330
So planar sharp 3SAT
is sharp P complete.

288
00:17:53,360 --> 00:17:56,200
I'd like to pretend that there's
some debate within the sharp P

289
00:17:56,200 --> 00:18:00,640
community about whether the
sharp P are here or here.

290
00:18:00,640 --> 00:18:03,580
I kind of prefer it here,
but I've seen it over here,

291
00:18:03,580 --> 00:18:04,240
so you know.

292
00:18:07,650 --> 00:18:10,140
I don't think I've even
seen that mentioned but I'm

293
00:18:10,140 --> 00:18:12,530
sure it's out there somewhere.

294
00:18:12,530 --> 00:18:16,550
So let's flip through our
other planar hardness proofs.

295
00:18:16,550 --> 00:18:19,920
This is planar monotone
rectilinear 3SAT.

296
00:18:19,920 --> 00:18:21,790
Mainly the monotone aspect.

297
00:18:21,790 --> 00:18:24,170
We wanted there to
be all the variables

298
00:18:24,170 --> 00:18:26,440
to be positive or
negative in every clause.

299
00:18:26,440 --> 00:18:29,820
And so we had this trick
for forcing this thing

300
00:18:29,820 --> 00:18:31,070
to be not equal to this thing.

301
00:18:31,070 --> 00:18:34,790
Basically copying the variable
with a flip in its truths.

302
00:18:34,790 --> 00:18:36,610
But again everything
here is forced.

303
00:18:36,610 --> 00:18:38,662
The variables we
make are guaranteed

304
00:18:38,662 --> 00:18:40,620
to be copies or indications
of other variables.

305
00:18:40,620 --> 00:18:42,550
So we preserve
number of solutions.

306
00:18:42,550 --> 00:18:43,790
So this is hard too.

307
00:19:00,260 --> 00:19:03,060
OK, what else?

308
00:19:03,060 --> 00:19:05,210
I think we can-- I
just made this up

309
00:19:05,210 --> 00:19:07,870
--but we can add the
dash three as usual

310
00:19:07,870 --> 00:19:10,000
by replacing each variable
with little cycle.

311
00:19:12,520 --> 00:19:14,530
What about planar 1-in-3 SAT?

312
00:19:14,530 --> 00:19:17,240
So we had, we actually had
two ways to prove this.

313
00:19:17,240 --> 00:19:23,180
This is one of the
reductions and we

314
00:19:23,180 --> 00:19:26,960
check that this set
of SAT clauses, these

315
00:19:26,960 --> 00:19:29,477
are the SAT clauses implemented
this 1-in-3 SAT clause.

316
00:19:29,477 --> 00:19:32,060
And I stared at this for a while
and its kind of hard to tell.

317
00:19:32,060 --> 00:19:34,420
So I just wrote a program
to enumerate all cases

318
00:19:34,420 --> 00:19:36,970
and found there's
exactly one case where

319
00:19:36,970 --> 00:19:39,790
this is not parsimonious.

320
00:19:39,790 --> 00:19:45,760
And that's when this is
false and these two are true.

321
00:19:45,760 --> 00:19:47,900
And because of these
negations you can either

322
00:19:47,900 --> 00:19:49,400
solve the internal
things like this,

323
00:19:49,400 --> 00:19:51,380
or you could flip all
of the internal nodes

324
00:19:51,380 --> 00:19:53,550
and that will also be satisfied.

325
00:19:53,550 --> 00:19:57,600
Now this is bad news, because
all the other cases there

326
00:19:57,600 --> 00:20:00,279
is a unique solution over here.

327
00:20:00,279 --> 00:20:02,320
But in this case there
are exactly two solutions.

328
00:20:02,320 --> 00:20:03,945
If it was two everywhere
we'd be happy,

329
00:20:03,945 --> 00:20:05,500
that would be twomonious.

330
00:20:05,500 --> 00:20:07,540
If it was one everywhere
we'd be happier, happy

331
00:20:07,540 --> 00:20:08,910
that would be parsimonious.

332
00:20:08,910 --> 00:20:11,210
But because it's a
mixture of one and two

333
00:20:11,210 --> 00:20:12,975
we have approximately
preserved the counts

334
00:20:12,975 --> 00:20:13,980
with a factor of two.

335
00:20:13,980 --> 00:20:15,890
But that's not good
enough for sharp P,

336
00:20:15,890 --> 00:20:19,240
we need exact preservation.

337
00:20:19,240 --> 00:20:20,960
So this is no good.

338
00:20:20,960 --> 00:20:23,200
Luckily we had another
proof which was actually

339
00:20:23,200 --> 00:20:27,400
a stronger result, Planar
Positive Rectilinear 1-in-3SAT.

340
00:20:27,400 --> 00:20:29,280
This was a version
with no negation,

341
00:20:29,280 --> 00:20:32,580
and this one does work.

342
00:20:32,580 --> 00:20:36,200
There's first of all
this, the not equal gadget

343
00:20:36,200 --> 00:20:37,220
and the equal gadget.

344
00:20:37,220 --> 00:20:40,850
I don't want to go through them,
but again A forced to be zero,

345
00:20:40,850 --> 00:20:43,780
C was forced to be one,
which forces B and D to be

346
00:20:43,780 --> 00:20:45,250
zero in this picture.

347
00:20:45,250 --> 00:20:48,930
So all is good.

348
00:20:48,930 --> 00:20:50,550
And again parsimonious there.

349
00:20:50,550 --> 00:20:53,400
And then this one was too
complicated to think about,

350
00:20:53,400 --> 00:20:55,413
so I again wrote a
program to check-- try

351
00:20:55,413 --> 00:20:59,120
all the cases and
every choice of XYZ

352
00:20:59,120 --> 00:21:02,040
over here that's satisfied,
this is a reduction from 3SAT.

353
00:21:02,040 --> 00:21:03,870
So if at least one
of these is true,

354
00:21:03,870 --> 00:21:06,420
there will be exactly
one solution over here.

355
00:21:06,420 --> 00:21:09,555
And just as before after--
if zero of them are true,

356
00:21:09,555 --> 00:21:10,930
then they'll be
no solution here.

357
00:21:10,930 --> 00:21:13,400
That we already knew.

358
00:21:13,400 --> 00:21:14,860
So good news.

359
00:21:14,860 --> 00:21:17,110
I should check whether that's
mentioned in their paper

360
00:21:17,110 --> 00:21:20,830
but it proves planar positive
rectilinear 1-in-3SAT

361
00:21:20,830 --> 00:21:22,270
is sharp P complete.

362
00:21:42,460 --> 00:21:43,720
Sharp go here, here.

363
00:21:46,260 --> 00:21:48,470
OK, so lots of fun results.

364
00:21:48,470 --> 00:21:50,680
We get a lot of results
just from-- by looking

365
00:21:50,680 --> 00:21:51,360
at old proofs.

366
00:21:51,360 --> 00:21:54,930
Now they're not
all going to work,

367
00:21:54,930 --> 00:21:57,650
but I have one more
example that does work.

368
00:21:57,650 --> 00:22:00,410
Shakashaka, remember the puzzle?

369
00:22:00,410 --> 00:22:01,850
It's a Nikoli puzzle.

370
00:22:01,850 --> 00:22:03,800
Every square,
every white squared

371
00:22:03,800 --> 00:22:07,250
can be filled in with a black
thing, but adjacent to it too,

372
00:22:07,250 --> 00:22:09,870
there should be
exactly two of those.

373
00:22:09,870 --> 00:22:12,050
And you want all of the
resulting white regions

374
00:22:12,050 --> 00:22:14,450
to be rectangles
possibly rotated.

375
00:22:14,450 --> 00:22:19,520
And we had this reduction
from planar 3SAT,

376
00:22:19,520 --> 00:22:22,480
and this basically,
this type of wire.

377
00:22:22,480 --> 00:22:26,674
And there's exactly two
ways to solve a wire.

378
00:22:26,674 --> 00:22:27,840
One for true, one for false.

379
00:22:27,840 --> 00:22:29,423
So once you know
what the variable is,

380
00:22:29,423 --> 00:22:31,240
you're forced what
to do, there is also

381
00:22:31,240 --> 00:22:33,700
a parity-shifting gadget
and splits and turns.

382
00:22:33,700 --> 00:22:36,530
But again, exactly two
ways to solve everything.

383
00:22:36,530 --> 00:22:39,300
So parsimonious, and then
the clause everything

384
00:22:39,300 --> 00:22:41,260
was basically
forced just-- you're

385
00:22:41,260 --> 00:22:44,100
forced whether to have
these square diamonds.

386
00:22:44,100 --> 00:22:45,700
And you just
eliminated the one case

387
00:22:45,700 --> 00:22:47,440
where the clause
is not satisfied.

388
00:22:47,440 --> 00:22:50,550
So there's really no flexibility
here, one way to solve it.

389
00:22:50,550 --> 00:22:52,310
And so it's a
parsimonious reduction

390
00:22:52,310 --> 00:22:54,450
and indeed in the
paper we mentioned

391
00:22:54,450 --> 00:22:57,330
this implies sharp P
completeness of counting

392
00:22:57,330 --> 00:22:59,770
Shakashaka solutions.

393
00:22:59,770 --> 00:23:02,010
Cool.

394
00:23:02,010 --> 00:23:04,900
Here's an example
that doesn't work.

395
00:23:04,900 --> 00:23:07,770
A little different,
Hamiltonicity or I

396
00:23:07,770 --> 00:23:11,130
guess I want to count the
number of Hamiltonian cycles.

397
00:23:11,130 --> 00:23:16,510
The natural counting
version of Hamiltonicity.

398
00:23:16,510 --> 00:23:19,080
So we had two proofs for
this, neither of them

399
00:23:19,080 --> 00:23:22,370
work in a sharp P sense.

400
00:23:22,370 --> 00:23:24,720
This one, remember
the idea was that you

401
00:23:24,720 --> 00:23:28,980
would traverse back and
forth one way or the other

402
00:23:28,980 --> 00:23:30,295
to get all of these nodes.

403
00:23:30,295 --> 00:23:31,670
That was a variable,
that's fine.

404
00:23:31,670 --> 00:23:33,425
There're exactly
two ways to do that.

405
00:23:33,425 --> 00:23:35,700
But then the clause,
the clause had

406
00:23:35,700 --> 00:23:39,290
to be satisfied by at least
one of the three variables.

407
00:23:39,290 --> 00:23:41,220
And if it's satisfied
for example,

408
00:23:41,220 --> 00:23:42,920
by all three variables,
then it could

409
00:23:42,920 --> 00:23:44,650
be this node is
picked up like this,

410
00:23:44,650 --> 00:23:46,934
or the node could be
picked up this way,

411
00:23:46,934 --> 00:23:48,600
or the node could be
picked up this way.

412
00:23:48,600 --> 00:23:50,700
So there are three
different solutions even

413
00:23:50,700 --> 00:23:53,130
for one fixed
variable assignment.

414
00:23:53,130 --> 00:23:55,370
So that's bad, we're
not allowed to do that.

415
00:23:55,370 --> 00:23:57,860
It'd be fine if every one was
three, but some will be one,

416
00:23:57,860 --> 00:23:59,870
some will be two,
some will be three.

417
00:23:59,870 --> 00:24:01,786
That's going to be some
weird product of those

418
00:24:01,786 --> 00:24:03,200
over the clauses.

419
00:24:03,200 --> 00:24:04,770
So that doesn't work.

420
00:24:04,770 --> 00:24:06,075
We had this other proof.

421
00:24:09,850 --> 00:24:13,430
This was a notation
for this gadget which

422
00:24:13,430 --> 00:24:16,610
forced either this directed
edge or this directed edge

423
00:24:16,610 --> 00:24:20,910
to be used, but not both.

424
00:24:20,910 --> 00:24:22,410
It's and x or.

425
00:24:22,410 --> 00:24:24,390
So that's, remember
what these things meant

426
00:24:24,390 --> 00:24:26,990
and that we have the
variable true or false.

427
00:24:26,990 --> 00:24:28,740
And then we connected
them to the clauses,

428
00:24:28,740 --> 00:24:31,860
then separately we
had a crossover.

429
00:24:31,860 --> 00:24:34,940
But the trouble
is in the clauses,

430
00:24:34,940 --> 00:24:38,810
because again, the idea was that
the variable chose this guy,

431
00:24:38,810 --> 00:24:40,500
this one was forbidden.

432
00:24:40,500 --> 00:24:42,910
That's actually the
good case I think.

433
00:24:42,910 --> 00:24:46,390
If the variable chose
this, chose this one,

434
00:24:46,390 --> 00:24:48,340
then this one must be included.

435
00:24:48,340 --> 00:24:52,661
That's bad news, if you
followed this, and then this,

436
00:24:52,661 --> 00:24:54,910
and then this, then you cut
off this part of the graph

437
00:24:54,910 --> 00:24:56,980
and you don't get one
Hamiltonian cycle.

438
00:24:56,980 --> 00:25:02,050
You want at least one variable
to allow you to go left here

439
00:25:02,050 --> 00:25:05,230
and then you can go and grab
all this stuff and come back.

440
00:25:05,230 --> 00:25:08,000
But again if multiple
variables satisfy this thing,

441
00:25:08,000 --> 00:25:10,800
any one of them could
grab the left rectangle.

442
00:25:10,800 --> 00:25:14,115
And so they get multiple
solutions, not parsimonious.

443
00:25:16,630 --> 00:25:18,990
But parts of this
proof are useful

444
00:25:18,990 --> 00:25:22,240
and they are used to make
a parsimonious proof.

445
00:25:22,240 --> 00:25:25,550
So the part that was useful is
this x or gadget, and the way

446
00:25:25,550 --> 00:25:27,200
to implement crossovers.

447
00:25:27,200 --> 00:25:32,390
So just remember that
you can build x ors,

448
00:25:32,390 --> 00:25:35,670
and that you can cross them
over using a bunch of x ors.

449
00:25:35,670 --> 00:25:38,610
So only x or connections,
these notations,

450
00:25:38,610 --> 00:25:41,010
can be crossed in this view.

451
00:25:41,010 --> 00:25:42,870
We're going to
build more gadgets

452
00:25:42,870 --> 00:25:46,195
and this is a proof
by Sato in 2002.

453
00:25:46,195 --> 00:25:50,120
It was a bachelor's
thesis actually in Japan.

454
00:25:50,120 --> 00:25:53,199
And so here you see
redrawn the x or gadget.

455
00:25:53,199 --> 00:25:54,990
Here it's going to be
for undirected graph,

456
00:25:54,990 --> 00:25:58,290
same structure works.

457
00:25:58,290 --> 00:26:01,110
And this is do the
crossover using x or.

458
00:26:01,110 --> 00:26:05,710
So he's denoting x or's as this
big X connected to two things.

459
00:26:05,710 --> 00:26:09,270
Now given that we can
build an or gadget, which

460
00:26:09,270 --> 00:26:11,880
says that either we use this
edge or we use this edge,

461
00:26:11,880 --> 00:26:14,400
or both.

462
00:26:14,400 --> 00:26:17,450
Here we're not using an x
or, but this is the graph.

463
00:26:17,450 --> 00:26:21,640
And the key is this has
to be done uniquely.

464
00:26:21,640 --> 00:26:24,330
That's in particular
the point of these dots.

465
00:26:24,330 --> 00:26:27,450
This looks asymmetric,
it's kind of weird.

466
00:26:27,450 --> 00:26:33,630
For example, if they're both
in, then this guy can do this,

467
00:26:33,630 --> 00:26:35,890
and this guy can do that.

468
00:26:35,890 --> 00:26:36,970
But it's not symmetric.

469
00:26:36,970 --> 00:26:40,594
You couldn't flip-- this guy
can't grab these points, that

470
00:26:40,594 --> 00:26:42,510
would be a second solution
which would be bad,

471
00:26:42,510 --> 00:26:43,850
but we missed these points.

472
00:26:43,850 --> 00:26:47,370
So this guy has to stay
down here if he's in at all.

473
00:26:47,370 --> 00:26:49,250
And then this guy's
the only one who

474
00:26:49,250 --> 00:26:52,120
can grab those extra points.

475
00:26:52,120 --> 00:26:57,565
Or if just the top guy's
in then you do this.

476
00:27:00,320 --> 00:27:06,080
And if just the bottom guy's
in I think it's symmetric.

477
00:27:06,080 --> 00:27:10,040
That is symmetric,
and it's unique.

478
00:27:10,040 --> 00:27:11,350
Good.

479
00:27:11,350 --> 00:27:16,130
If there's an implication-- so
this says if this edge is in,

480
00:27:16,130 --> 00:27:20,210
then that edge must be
in the Hamiltonian cycle

481
00:27:20,210 --> 00:27:23,310
and this is
essentially by copying.

482
00:27:23,310 --> 00:27:25,290
And we just have to
grab an extra edge

483
00:27:25,290 --> 00:27:28,780
and add this little extra
thing just for copying value.

484
00:27:28,780 --> 00:27:33,330
So if this one is in, then
this edge must not be used,

485
00:27:33,330 --> 00:27:36,147
which means this edge if
it's used must go straight.

486
00:27:36,147 --> 00:27:37,480
In particular, this is not used.

487
00:27:37,480 --> 00:27:39,520
And we have an or that
means that this one must

488
00:27:39,520 --> 00:27:41,930
be forced by a property of or.

489
00:27:41,930 --> 00:27:47,950
On the other hand, if this is
not set, this one must be used.

490
00:27:47,950 --> 00:27:51,150
So I guess this must be
an edge that was already

491
00:27:51,150 --> 00:27:53,640
going to be used for something.

492
00:27:53,640 --> 00:27:56,350
So that edge is just going
to get diverted down here,

493
00:27:56,350 --> 00:27:59,390
and then the or doesn't
constrain us at all.

494
00:27:59,390 --> 00:28:03,390
Because zero or one, this one
is one so the or is happy.

495
00:28:03,390 --> 00:28:05,200
So that's an implication.

496
00:28:05,200 --> 00:28:08,666
It's going to be a little more
subtle how we combine these.

497
00:28:08,666 --> 00:28:12,700
This is the tricky gadget.

498
00:28:12,700 --> 00:28:15,390
I sort of understand
it, but it has

499
00:28:15,390 --> 00:28:17,110
a lot of details to
check, especially

500
00:28:17,110 --> 00:28:18,770
on the uniqueness front.

501
00:28:18,770 --> 00:28:20,560
But this is a three-way
or which we're

502
00:28:20,560 --> 00:28:24,480
using for clause, SAT
clause, 3SAT clause.

503
00:28:24,480 --> 00:28:27,420
We want at least one
of these three edges

504
00:28:27,420 --> 00:28:29,230
to be in the Hamiltonian cycle.

505
00:28:29,230 --> 00:28:34,570
And so here we use x ors, ors,
implications, and more x ors.

506
00:28:34,570 --> 00:28:36,260
I'll show you the
intended solution,

507
00:28:36,260 --> 00:28:39,670
assuming I can remember it.

508
00:28:39,670 --> 00:28:43,940
So let's say for example, this
edge is in the Hamiltonian

509
00:28:43,940 --> 00:28:45,750
cycle.

510
00:28:45,750 --> 00:28:54,130
Then we're going to do
something like go over here,

511
00:28:54,130 --> 00:28:57,250
come back around like this.

512
00:28:57,250 --> 00:28:57,980
And then

513
00:28:57,980 --> 00:29:00,584
AUDIENCE: Did you just
violate the x or already?

514
00:29:00,584 --> 00:29:01,820
PROFESSOR: This x or?

515
00:29:01,820 --> 00:29:02,565
AUDIENCE: Yeah.

516
00:29:02,565 --> 00:29:03,690
PROFESSOR: No, I went here.

517
00:29:03,690 --> 00:29:04,610
AUDIENCE: And then you went up.

518
00:29:04,610 --> 00:29:06,150
PROFESSOR: Oh, then
I went up, good.

519
00:29:06,150 --> 00:29:08,970
So actually I have to
do this and around.

520
00:29:08,970 --> 00:29:11,230
Yes, so all these constraints
are thrown in, basically

521
00:29:11,230 --> 00:29:12,271
to force it to be unique.

522
00:29:12,271 --> 00:29:14,890
Without them you could--
still it would still work,

523
00:29:14,890 --> 00:29:17,620
but it wouldn't be parsimonious.

524
00:29:17,620 --> 00:29:19,620
OK, let's see, this--

525
00:29:19,620 --> 00:29:20,620
AUDIENCE: In the middle.

526
00:29:20,620 --> 00:29:22,700
PROFESSOR: --doesn't
constrain me.

527
00:29:22,700 --> 00:29:23,700
This one?

528
00:29:23,700 --> 00:29:24,846
AUDIENCE: Yeah.

529
00:29:24,846 --> 00:29:27,270
PROFESSOR: Good, go up there.

530
00:29:27,270 --> 00:29:32,650
So I did exactly one of those
and then I need to grab these.

531
00:29:32,650 --> 00:29:34,060
OK, so that's one picture.

532
00:29:34,060 --> 00:29:37,300
I think it's not
totally symmetric.

533
00:29:37,300 --> 00:29:39,390
Again, so you have to
check all three of them.

534
00:29:39,390 --> 00:29:42,580
And you have to check for
example, It's not so hard.

535
00:29:42,580 --> 00:29:45,270
Like if this guy was also in,
I could have just gone here

536
00:29:45,270 --> 00:29:47,610
and this guy would
pick up those nodes.

537
00:29:47,610 --> 00:29:50,900
So as long as, in general, as
long as at least one of them

538
00:29:50,900 --> 00:29:53,450
is on you're OK.

539
00:29:53,450 --> 00:29:55,496
And furthermore,
if they're all on,

540
00:29:55,496 --> 00:29:57,120
there's still a unique
way to solve it.

541
00:29:57,120 --> 00:29:58,430
And I'm not going
to go through that.

542
00:29:58,430 --> 00:30:00,340
But it's thanks to all
of these constraints

543
00:30:00,340 --> 00:30:04,110
they're cutting out
multiple solutions.

544
00:30:04,110 --> 00:30:08,210
OK, so now we just had
to put these together,

545
00:30:08,210 --> 00:30:09,830
this is pretty easy.

546
00:30:09,830 --> 00:30:11,410
It's pretty much
like the old proof.

547
00:30:11,410 --> 00:30:14,010
Again, we have-- we
represent variables

548
00:30:14,010 --> 00:30:18,250
by having these double edges
in a Hamiltonian cycle.

549
00:30:18,250 --> 00:30:21,580
You're going to choose
one or the other.

550
00:30:21,580 --> 00:30:24,200
And then we have this exclusive
or, forcing y and y bar

551
00:30:24,200 --> 00:30:25,950
to choose opposite choices.

552
00:30:25,950 --> 00:30:27,950
And then there are these
exclusive words to say,

553
00:30:27,950 --> 00:30:29,908
if you chose that one,
then you can't choose it

554
00:30:29,908 --> 00:30:31,222
for this clause.

555
00:30:31,222 --> 00:30:32,930
And then the clauses
are just represented

556
00:30:32,930 --> 00:30:34,190
by those three-way ors.

557
00:30:34,190 --> 00:30:36,190
So this overall
structure is the same.

558
00:30:36,190 --> 00:30:38,340
The crossovers
are done as before

559
00:30:38,340 --> 00:30:41,150
and it's really these
gadgets that have changed.

560
00:30:41,150 --> 00:30:43,420
And they're complicated,
but it's parsimonious.

561
00:30:43,420 --> 00:30:45,700
So with a little more work.

562
00:30:45,700 --> 00:30:48,630
We get the number of
Hamiltonian cycles

563
00:30:48,630 --> 00:30:54,400
in planar max degree three
graphs is sharp P complete.

564
00:31:16,920 --> 00:31:20,030
So that's nice.

565
00:31:20,030 --> 00:31:24,390
And from that, we
get Slitherlink.

566
00:31:24,390 --> 00:31:27,684
So this is-- I've sort of been
hiding these facts from you.

567
00:31:27,684 --> 00:31:29,350
When we originally
covered these proofs,

568
00:31:29,350 --> 00:31:33,240
these papers actually talk
about-- This one doesn't quite

569
00:31:33,240 --> 00:31:35,640
talk about sharp P, but it
is also sharp P complete.

570
00:31:35,640 --> 00:31:37,960
Counting the number of
solutions to Slitherlink

571
00:31:37,960 --> 00:31:40,207
is sharp P complete.

572
00:31:40,207 --> 00:31:41,040
This was the puzzle.

573
00:31:41,040 --> 00:31:45,530
Again you have that number of
adjacent edges-- each number.

574
00:31:45,530 --> 00:31:49,740
And the proof was a
reduction from planar max

575
00:31:49,740 --> 00:31:52,810
degree three Hamiltonian cycle.

576
00:31:52,810 --> 00:31:54,740
And at the time I
said, oh you could just

577
00:31:54,740 --> 00:31:56,420
assume it's a grid graph.

578
00:31:56,420 --> 00:31:59,880
And then you just need
the required gadget,

579
00:31:59,880 --> 00:32:01,550
which is the b part.

580
00:32:01,550 --> 00:32:02,970
Just need this gadget.

581
00:32:02,970 --> 00:32:05,010
This was a gadget because
it had these ones.

582
00:32:05,010 --> 00:32:08,045
It meant it had to be
traversed like a regular vertex

583
00:32:08,045 --> 00:32:10,380
in Hamiltonian cycle
and it turns out

584
00:32:10,380 --> 00:32:13,260
there was a way to traverse
it straight or with returns.

585
00:32:13,260 --> 00:32:15,610
And then we could block
off edges wherever there

586
00:32:15,610 --> 00:32:16,860
wasn't supposed to be an edge.

587
00:32:16,860 --> 00:32:19,390
And so if you're reducing from
Hamiltonicity and grid graphs

588
00:32:19,390 --> 00:32:22,720
that was the whole
proof and we were happy.

589
00:32:22,720 --> 00:32:25,280
Now we don't know whether
Hamiltonicity and grid

590
00:32:25,280 --> 00:32:26,586
graphs is sharp P complete.

591
00:32:30,796 --> 00:32:32,170
To prove that we
would need to be

592
00:32:32,170 --> 00:32:34,040
able to put bipartite
in here, and I

593
00:32:34,040 --> 00:32:35,610
don't know of a proof of that.

594
00:32:35,610 --> 00:32:37,390
Good open problem.

595
00:32:37,390 --> 00:32:40,450
So this was the reason that they
have this other gadget, which

596
00:32:40,450 --> 00:32:45,140
was to make Hamiltonian-- to
make these white vertices that

597
00:32:45,140 --> 00:32:46,920
don't have to be traversed.

598
00:32:46,920 --> 00:32:49,390
They're just implementing
an edge basically.

599
00:32:49,390 --> 00:32:51,600
So just the black vertices
have to be traversed.

600
00:32:51,600 --> 00:32:54,660
We needed that for drawing
things on the grid.

601
00:32:54,660 --> 00:32:57,790
But if you just don't put in
the ones and add in these zeros

602
00:32:57,790 --> 00:33:01,320
again you can traverse
it or not all.

603
00:33:01,320 --> 00:33:03,340
And this one, as
you can read here,

604
00:33:03,340 --> 00:33:06,400
says this would be
bad to have happen.

605
00:33:06,400 --> 00:33:08,150
In fact you can rule
it out, it will never

606
00:33:08,150 --> 00:33:10,390
happen in the
situations we want.

607
00:33:10,390 --> 00:33:13,280
Because the white vertices
only have two neighbors

608
00:33:13,280 --> 00:33:15,010
that are traversable.

609
00:33:18,310 --> 00:33:22,240
Cool, and then furthermore, all
of these solutions are unique.

610
00:33:22,240 --> 00:33:24,540
There is exactly one
way to go straight.

611
00:33:24,540 --> 00:33:26,830
There's exactly one way to
turn right, exactly one way

612
00:33:26,830 --> 00:33:27,980
to turn left.

613
00:33:27,980 --> 00:33:30,490
That's the subtle thing that
was not revealed before.

614
00:33:30,490 --> 00:33:34,240
But if you stare at it long
enough you will be convinced.

615
00:33:34,240 --> 00:33:36,590
So this is a
parsimonious reduction

616
00:33:36,590 --> 00:33:38,940
from planar max degree
three Hamiltonian

617
00:33:38,940 --> 00:33:40,840
cycle to Slitherlink.

618
00:33:40,840 --> 00:33:43,870
So counting solutions
in Slitherlink is hard.

619
00:33:43,870 --> 00:33:48,860
That was the fully
worked out example.

620
00:33:48,860 --> 00:33:50,412
Any questions?

621
00:33:50,412 --> 00:33:50,912
Yeah.

622
00:33:50,912 --> 00:33:53,200
AUDIENCE: So are there
examples of problems in P

623
00:33:53,200 --> 00:33:56,380
who's counting versions
are sharp P complete?

624
00:33:56,380 --> 00:33:57,060
PROFESSOR: Yes.

625
00:33:57,060 --> 00:34:00,401
And that will be the next topic.

626
00:34:00,401 --> 00:34:02,650
Well it's going to be a
little while til we get there.

627
00:34:02,650 --> 00:34:04,440
But I'm going to proof
things and then we

628
00:34:04,440 --> 00:34:06,390
will get to an answer,
but the answer is yes.

629
00:34:10,531 --> 00:34:11,322
AUDIENCE: Question.

630
00:34:11,322 --> 00:34:12,530
PROFESSOR: Yeah.

631
00:34:12,530 --> 00:34:14,874
AUDIENCE: So, for if
we wanted-- if we're

632
00:34:14,874 --> 00:34:16,540
like thinking about
a problem that we're

633
00:34:16,540 --> 00:34:18,329
trying to prove something
P-hard and we start thinking

634
00:34:18,329 --> 00:34:20,912
maybe stop, we would just show
it's in P by finding algorithm.

635
00:34:20,912 --> 00:34:25,500
Is there a nice way to show
that our problem is not P hard?

636
00:34:25,500 --> 00:34:28,900
PROFESSOR: Well you usually
would say that sharp,

637
00:34:28,900 --> 00:34:31,300
that problem is NP.

638
00:34:31,300 --> 00:34:33,830
You find a polynomial
counting algorithm.

639
00:34:33,830 --> 00:34:37,190
And they're lot's of examples of
polynomial counting algorithms

640
00:34:37,190 --> 00:34:38,840
especially on like trees.

641
00:34:38,840 --> 00:34:40,771
Typical thing is
your dynamic program.

642
00:34:40,771 --> 00:34:42,520
So like maybe you want
to know-- let's say

643
00:34:42,520 --> 00:34:44,179
you have a rooted binary
tree and for each node

644
00:34:44,179 --> 00:34:45,889
you could flip it
this way or not.

645
00:34:45,889 --> 00:34:48,250
How many different ways
are there to do that?

646
00:34:48,250 --> 00:34:50,699
And maybe have some
constraints on how that's done.

647
00:34:50,699 --> 00:34:53,122
Then you just try it
flipped, and try it not.

648
00:34:53,122 --> 00:34:54,580
You do dynamic
programming and then

649
00:34:54,580 --> 00:34:57,937
you multiply the two
solution sizes together,

650
00:34:57,937 --> 00:34:59,520
and you get the
overall solution size.

651
00:34:59,520 --> 00:35:01,380
So you basically
do combinatorics

652
00:35:01,380 --> 00:35:03,130
and if there's
independent choices you

653
00:35:03,130 --> 00:35:06,540
multiply, if they're
opposing choices you add,

654
00:35:06,540 --> 00:35:07,410
that kind of thing.

655
00:35:07,410 --> 00:35:10,070
And from that you get polynomial
time counting algorithms.

656
00:35:10,070 --> 00:35:14,040
In tree-like things that
often works inbound treewidth.

657
00:35:14,040 --> 00:35:16,562
AUDIENCE: Do you know that NP
hard problems who's counting

658
00:35:16,562 --> 00:35:19,567
problems are not sharp P hard?

659
00:35:19,567 --> 00:35:21,108
I guess this technique
wouldn't work.

660
00:35:21,108 --> 00:35:23,890
PROFESSOR: I would say
generally most problems

661
00:35:23,890 --> 00:35:26,190
that are hard to decide
are hard to count.

662
00:35:26,190 --> 00:35:29,120
And where NP hard
implies sharp P hard.

663
00:35:29,120 --> 00:35:33,329
I don't think there's a
hard theorem in that there's

664
00:35:33,329 --> 00:35:35,620
nothing that really says--
meta-theorem that says that,

665
00:35:35,620 --> 00:35:36,578
but that's the feeling.

666
00:35:39,142 --> 00:35:40,600
It'd be nice, then
we wouldn't have

667
00:35:40,600 --> 00:35:42,070
to do all the parsimonious work.

668
00:35:45,700 --> 00:35:50,350
All right so it's time for a
little bit of linear algebra.

669
00:35:54,410 --> 00:35:57,270
Let me remind you, I
guess linear algebra's not

670
00:35:57,270 --> 00:35:59,350
a pre-req for this class,
but probably you've

671
00:35:59,350 --> 00:36:01,489
seen the determinant
of a matrix and use,

672
00:36:01,489 --> 00:36:03,530
if it's zero then it's
non-invertible blah, blah,

673
00:36:03,530 --> 00:36:04,784
blah.

674
00:36:04,784 --> 00:36:06,200
Let me remind you
of a definition.

675
00:36:11,980 --> 00:36:13,640
And we rarely use
matrix notation.

676
00:36:13,640 --> 00:36:17,540
So let me remind you
of the usual one.

677
00:36:17,540 --> 00:36:19,050
N by n, square matrix.

678
00:36:19,050 --> 00:36:21,130
This is a polynomial
time problem.

679
00:36:21,130 --> 00:36:25,330
It is-- but I'm going to define
it in an exponential way.

680
00:36:25,330 --> 00:36:28,410
But you probably know a
polynomial time algorithm.

681
00:36:28,410 --> 00:36:31,937
This is not an algorithms class
so you don't need to know it.

682
00:36:34,760 --> 00:36:37,010
But it's based on Gaussian
elimination, the usual one.

683
00:36:56,370 --> 00:37:00,790
So you look at all
permutation matrices,

684
00:37:00,790 --> 00:37:02,580
all n by n permutation
matrices which

685
00:37:02,580 --> 00:37:05,860
you can think of as a
permutation pi on the numbers 1

686
00:37:05,860 --> 00:37:06,880
through n.

687
00:37:06,880 --> 00:37:09,780
And you look at i
comma pi event, that

688
00:37:09,780 --> 00:37:11,410
defines the permutation matrix.

689
00:37:11,410 --> 00:37:14,500
You take the product
of the matrix

690
00:37:14,500 --> 00:37:20,140
values-- if you superimpose
the permutation matrix on that,

691
00:37:20,140 --> 00:37:23,995
on the given matrix A.

692
00:37:23,995 --> 00:37:28,040
You take that product
you possibly negate it

693
00:37:28,040 --> 00:37:30,690
if the sign of your
permutation was negative,

694
00:37:30,690 --> 00:37:35,960
if it does an even number-- an
odd number of transpositions

695
00:37:35,960 --> 00:37:37,990
then this will be negative.

696
00:37:37,990 --> 00:37:40,780
Otherwise, it'd be positive
and you add those up.

697
00:37:40,780 --> 00:37:42,840
So of course, an exponential
number permutations

698
00:37:42,840 --> 00:37:45,730
you wouldn't want to do this as
an algorithm but turns out it

699
00:37:45,730 --> 00:37:49,745
can be done in polynomial time.

700
00:37:49,745 --> 00:37:51,120
The reason for
talking about this

701
00:37:51,120 --> 00:37:56,690
is by analogy I want the
notion of permanent, of an n

702
00:37:56,690 --> 00:37:57,450
by n matrix.

703
00:37:57,450 --> 00:37:59,190
The same thing, but
with this removed.

704
00:38:02,370 --> 00:38:06,620
The determinant of A is the
sum over all permutations pi,

705
00:38:06,620 --> 00:38:13,900
of the product from i equals
one to n of ai, pi of i.

706
00:38:13,900 --> 00:38:15,870
Now this may not look
like a counting problem.

707
00:38:15,870 --> 00:38:17,900
It turns out it is a
counting problem, sort of,

708
00:38:17,900 --> 00:38:19,232
a weighted counting problem.

709
00:38:19,232 --> 00:38:21,315
We will get back to counting
problems in a moment.

710
00:38:21,315 --> 00:38:24,740
This Is related to the number
of perfect matchings in a graph.

711
00:38:24,740 --> 00:38:26,520
But at this point
it's just-- it's

712
00:38:26,520 --> 00:38:27,800
a quantity we want to compute.

713
00:38:27,800 --> 00:38:29,630
This is a function of a matrix.

714
00:38:29,630 --> 00:38:33,600
And computing this function
is sharp P complete.

715
00:38:33,600 --> 00:38:34,100
Yeah.

716
00:38:34,100 --> 00:38:37,260
AUDIENCE: Could it just be
sine minus 1 2e sine of pi?

717
00:38:40,276 --> 00:38:42,310
PROFESSOR: I don't
know, it depends.

718
00:38:42,310 --> 00:38:46,695
If you call the number of
transpositions mod two so then

719
00:38:46,695 --> 00:38:48,045
it's zero one.

720
00:38:48,045 --> 00:38:48,920
You know what I mean.

721
00:38:53,210 --> 00:38:56,040
All right.

722
00:38:56,040 --> 00:38:58,017
So the claim is, permanent
is sharp P complete.

723
00:38:58,017 --> 00:38:59,100
We're going to prove this.

724
00:38:59,100 --> 00:39:01,970
This was the original problem
proved sharp P complete.

725
00:39:01,970 --> 00:39:05,230
Well other than
sharp 3SAT I guess.

726
00:39:05,230 --> 00:39:05,780
Same paper.

727
00:39:09,680 --> 00:39:15,130
Great so let me give
you a little intuition

728
00:39:15,130 --> 00:39:16,740
of what the permanent is.

729
00:39:16,740 --> 00:39:20,496
We'd like a definition
that's not so algebraic.

730
00:39:20,496 --> 00:39:23,980
At least I would like one more
graph-theoretic would be nice.

731
00:39:23,980 --> 00:39:25,810
So here's what
we're going to do.

732
00:39:25,810 --> 00:39:30,120
We're going to convert A, our
matrix, into a weighted graph.

733
00:39:42,130 --> 00:39:49,520
And then let me go
to the other board.

734
00:40:00,580 --> 00:40:02,251
How do we convert
into a weighted graph,

735
00:40:02,251 --> 00:40:03,250
weighted directed graph?

736
00:40:03,250 --> 00:40:08,420
Well the weight from,
vertex I to vertex J is AIJ.

737
00:40:08,420 --> 00:40:10,950
The obvious transformation
if it's zero then

738
00:40:10,950 --> 00:40:13,450
there's not going to be an edge,
although it doesn't matter.

739
00:40:13,450 --> 00:40:15,325
You could leave the edge
in with weight zero,

740
00:40:15,325 --> 00:40:17,004
it will be the same.

741
00:40:17,004 --> 00:40:18,920
Because what we're
interested in, in the claim

742
00:40:18,920 --> 00:40:28,430
is the permanent of the
matrix equals the sum,

743
00:40:28,430 --> 00:40:42,217
of the product of edge
weights, over all cycle covers

744
00:40:42,217 --> 00:40:46,040
of the graph.

745
00:40:46,040 --> 00:40:49,240
OK, this is really just the same
thing, but a little bit easier

746
00:40:49,240 --> 00:40:50,400
to think about.

747
00:40:50,400 --> 00:40:53,330
So a cycle cover it's kind
of like a Hamiltonian cycle

748
00:40:53,330 --> 00:40:55,440
but there could be
multiple cycles.

749
00:40:55,440 --> 00:40:58,600
So at every vertex you
should enter and leave.

750
00:40:58,600 --> 00:41:01,430
And so you have sort of in
degree one and out degree one.

751
00:41:01,430 --> 00:41:04,980
So you end up decomposing the
graph into vertex destroyed

752
00:41:04,980 --> 00:41:08,940
cycles which hit every vertex.

753
00:41:08,940 --> 00:41:11,250
That's a cycle cover.

754
00:41:11,250 --> 00:41:14,190
Important note here, we do
have loops in the graph.

755
00:41:14,190 --> 00:41:17,140
We can't have loops
if aii is not zero.

756
00:41:17,140 --> 00:41:18,370
Then you have a loop.

757
00:41:18,370 --> 00:41:22,840
So the idea is to look
at every cycle cover

758
00:41:22,840 --> 00:41:25,630
and just take the product
of the edge weights

759
00:41:25,630 --> 00:41:28,550
that are edges in
the cycles and then

760
00:41:28,550 --> 00:41:30,270
add that up overall
cycle covers.

761
00:41:30,270 --> 00:41:32,210
So it's the same thing
if you stare at it

762
00:41:32,210 --> 00:41:34,135
long enough because
you're going from I.

763
00:41:34,135 --> 00:41:35,680
I mean this is
basically the cycle

764
00:41:35,680 --> 00:41:39,410
decomposition of the permutation
if you know permutation theory.

765
00:41:39,410 --> 00:41:40,910
So if you don't,
don't worry this is

766
00:41:40,910 --> 00:41:43,880
your definition of permanent.

767
00:41:43,880 --> 00:41:46,840
So we're going to prove this
problem, is sharp P complete?

768
00:41:55,360 --> 00:42:05,100
And we're going to prove
it by a C-monious reduction

769
00:42:05,100 --> 00:42:07,380
from sharp 3SAT.

770
00:42:10,718 --> 00:42:14,742
AUDIENCE: You said this is just
the original thing introducing

771
00:42:14,742 --> 00:42:15,242
that?

772
00:42:15,242 --> 00:42:16,530
PROFESSOR: Yes.

773
00:42:16,530 --> 00:42:20,130
AUDIENCE: Did they not call it--
you made up the term C-monious.

774
00:42:20,130 --> 00:42:22,010
What did they call it?

775
00:42:22,010 --> 00:42:27,260
PROFESSOR: I think they
just called it a reduction.

776
00:42:27,260 --> 00:42:29,820
I think pretty sure, they
just called it reduction.

777
00:42:29,820 --> 00:42:32,060
And for them-- and they
said at the beginning,

778
00:42:32,060 --> 00:42:34,990
--reduction means
multicall reduction.

779
00:42:34,990 --> 00:42:36,850
So they're thinking about that.

780
00:42:36,850 --> 00:42:40,190
But it turns out to be
a C-monious reduction.

781
00:42:40,190 --> 00:42:42,120
C will not be
literally constant,

782
00:42:42,120 --> 00:42:46,140
but it will be a function
of the problem sets.

783
00:42:46,140 --> 00:42:48,020
And this is the reduction.

784
00:42:48,020 --> 00:42:51,910
So as usual, we have a clause
gadget, and a variable gadget,

785
00:42:51,910 --> 00:42:53,770
and then there's
this shaded thing

786
00:42:53,770 --> 00:42:58,776
which is this matrix, which
you can think of as a graph.

787
00:42:58,776 --> 00:43:00,650
But in this case will
be easier to just think

788
00:43:00,650 --> 00:43:03,180
of as a black box matrix.

789
00:43:03,180 --> 00:43:07,990
OK all of the edges in these
pictures have weight one.

790
00:43:07,990 --> 00:43:09,580
And then these
edges are special,

791
00:43:09,580 --> 00:43:12,730
and here you have some
loops in the center.

792
00:43:12,730 --> 00:43:14,980
No one else has a loop.

793
00:43:14,980 --> 00:43:18,500
So the high-level
idea is if you're

794
00:43:18,500 --> 00:43:22,680
thinking of a cycle
cover in a vertex

795
00:43:22,680 --> 00:43:25,140
because you--
sorry, in a variable

796
00:43:25,140 --> 00:43:28,400
because you've got a vertex
here and a vertex here you

797
00:43:28,400 --> 00:43:29,650
have to cover them somehow.

798
00:43:29,650 --> 00:43:33,290
And the intent is that you
either cover this one this way,

799
00:43:33,290 --> 00:43:35,470
or you cover this
one that way, those

800
00:43:35,470 --> 00:43:38,620
would be the true and the false.

801
00:43:38,620 --> 00:43:42,100
And then from the
clouds perspective

802
00:43:42,100 --> 00:43:46,031
we need to understand, so then
these things are connected.

803
00:43:46,031 --> 00:43:47,655
This thing would go
here and this thing

804
00:43:47,655 --> 00:43:50,440
would go here, and
generally connect variables

805
00:43:50,440 --> 00:43:53,530
to clauses in the obvious way.

806
00:43:53,530 --> 00:43:56,515
And in general, for every
occurrence of the positive form

807
00:43:56,515 --> 00:43:59,420
and the negative form, sorry
positive or negative form

808
00:43:59,420 --> 00:44:02,500
you'll have one of these blobs
that connects to the clause.

809
00:44:02,500 --> 00:44:04,920
So overall architecture
should be clear.

810
00:44:04,920 --> 00:44:07,100
What does this gadget do?

811
00:44:07,100 --> 00:44:13,550
It has some nifty properties
let me write them down.

812
00:44:13,550 --> 00:44:17,790
So this matrix is
called x in the paper.

813
00:44:17,790 --> 00:44:23,299
So first of all, permanent
of x equals zero.

814
00:44:23,299 --> 00:44:24,840
I'm just going to
state these, I mean

815
00:44:24,840 --> 00:44:27,120
you could check them by
doing the computation.

816
00:44:27,120 --> 00:44:32,090
So we're interested in cycle
covers whose products are not

817
00:44:32,090 --> 00:44:32,590
zero.

818
00:44:32,590 --> 00:44:34,850
Otherwise they don't
contribute to the sum.

819
00:44:34,850 --> 00:44:39,310
So I could also add in non-zero.

820
00:44:39,310 --> 00:44:41,860
Meaning the product is non-zero.

821
00:44:41,860 --> 00:44:46,840
OK so if we had a cycle cover
that just-- where the cycle

822
00:44:46,840 --> 00:44:50,040
cover just locally solved this
thing by traversing these four

823
00:44:50,040 --> 00:44:52,170
vertices all by themselves.

824
00:44:52,170 --> 00:44:56,162
Then that cycle would
have permanent zero.

825
00:44:56,162 --> 00:44:57,870
And then the permanent
of the whole cycle

826
00:44:57,870 --> 00:45:01,090
covers the product
of those things.

827
00:45:01,090 --> 00:45:02,780
And so the overall
thing would be zero.

828
00:45:02,780 --> 00:45:04,220
So if you look at
a nonzero cycle

829
00:45:04,220 --> 00:45:06,400
cover you can't just
leave these isolated,

830
00:45:06,400 --> 00:45:08,180
you have to visit them.

831
00:45:08,180 --> 00:45:11,384
You have to enter
them and leave them.

832
00:45:11,384 --> 00:45:13,800
Now the question is, where
could you enter and leave them?

833
00:45:13,800 --> 00:45:16,710
This is maybe not totally
clear from the drawing

834
00:45:16,710 --> 00:45:21,230
but, the intent is that
the first vertex in-- which

835
00:45:21,230 --> 00:45:24,980
corresponds to row one, column
one here, is the left side.

836
00:45:24,980 --> 00:45:29,580
And the column four, row
four is the right side.

837
00:45:29,580 --> 00:45:31,940
I claim that you have
to enter at one of those

838
00:45:31,940 --> 00:45:33,960
and leave at the other.

839
00:45:33,960 --> 00:45:36,040
Why is that true?

840
00:45:36,040 --> 00:45:37,040
For a couple reasons.

841
00:45:37,040 --> 00:45:42,950
One is that the permanent
of x with row and column

842
00:45:42,950 --> 00:45:51,552
one removed is zero,
and so is the permanent

843
00:45:51,552 --> 00:45:54,210
of x with row and
column four removed.

844
00:45:58,810 --> 00:46:04,660
OK vertex one and four
out of this x matrix

845
00:46:04,660 --> 00:46:06,650
are the only places you
could enter and leave.

846
00:46:06,650 --> 00:46:08,950
But it's possible you enter
and then immediately leave.

847
00:46:08,950 --> 00:46:10,842
So you just touch
the thing and leave.

848
00:46:10,842 --> 00:46:12,800
That would correspond to
leaving behind a three

849
00:46:12,800 --> 00:46:14,330
by three sub-matrix.

850
00:46:14,330 --> 00:46:16,070
Either by deleting
this row and column,

851
00:46:16,070 --> 00:46:18,236
or by deleting this row and
column if you just visit

852
00:46:18,236 --> 00:46:19,549
this vertex and leave.

853
00:46:19,549 --> 00:46:20,840
Those also have permanent zero.

854
00:46:20,840 --> 00:46:23,210
So if you're looking at
a nonzero cycle cover

855
00:46:23,210 --> 00:46:24,740
you can't do that.

856
00:46:24,740 --> 00:46:27,360
So together those mean
that you enter one of them

857
00:46:27,360 --> 00:46:29,530
and leave at the other.

858
00:46:29,530 --> 00:46:43,380
And furthermore, if you look
at the permanent of x with rows

859
00:46:43,380 --> 00:46:49,980
and columns one and four
removed, both removed,

860
00:46:49,980 --> 00:46:51,600
that's also zero.

861
00:46:51,600 --> 00:46:55,550
So the permanent of
this sub-matrix is zero

862
00:46:55,550 --> 00:46:58,360
and therefore you can't
just enter here, jump here,

863
00:46:58,360 --> 00:46:59,280
and leave.

864
00:46:59,280 --> 00:47:02,894
Which means finally you have
to traverse all four vertices.

865
00:47:02,894 --> 00:47:04,810
You enter at one of them,
traverse everything,

866
00:47:04,810 --> 00:47:06,470
and leave at the other.

867
00:47:06,470 --> 00:47:10,960
So basically this
is a forced edge.

868
00:47:10,960 --> 00:47:15,050
If you touch here you have to
then traverse and leave there,

869
00:47:15,050 --> 00:47:17,470
in any cycle cover.

870
00:47:17,470 --> 00:47:20,200
So we're used to seeing forced
edges in Hamiltonian cycle.

871
00:47:20,200 --> 00:47:22,430
This is sort of a
stronger form of it.

872
00:47:22,430 --> 00:47:26,300
That's cool now one catch.

873
00:47:26,300 --> 00:47:29,230
So if you do that, if you
enter let's say vertex one

874
00:47:29,230 --> 00:47:33,510
and leave at vertex four, you
will end up-- your contribution

875
00:47:33,510 --> 00:47:38,020
to the cycle will end up
being the permanent of x

876
00:47:38,020 --> 00:47:41,930
minus row one and column four.

877
00:47:41,930 --> 00:47:44,810
Or symmetrically
with four and one.

878
00:47:44,810 --> 00:47:49,740
You'd like this to be
one, but it's four.

879
00:47:49,740 --> 00:47:54,410
So there are four ways to
traverse the forced edge.

880
00:47:54,410 --> 00:47:56,910
But because it's
always four, or zero,

881
00:47:56,910 --> 00:47:58,520
and then it doesn't
contribute at all.

882
00:47:58,520 --> 00:48:01,410
It's always going to be four
so this will be a nice uniform

883
00:48:01,410 --> 00:48:03,862
blow up in our
number of solutions.

884
00:48:03,862 --> 00:48:05,320
C is not going to
be four, but it's

885
00:48:05,320 --> 00:48:08,400
going to be four to some power.

886
00:48:08,400 --> 00:48:10,310
It's going to be
four to the power,

887
00:48:10,310 --> 00:48:12,800
the number of those gadgets.

888
00:48:12,800 --> 00:48:14,247
So because you can
predict that, I

889
00:48:14,247 --> 00:48:16,830
mean in the reduction you know
exactly how many there it's not

890
00:48:16,830 --> 00:48:19,875
depend on the solution,
it's only dependent on how

891
00:48:19,875 --> 00:48:20,750
you built this thing.

892
00:48:20,750 --> 00:48:22,670
So at the end, we're
going to divide by four

893
00:48:22,670 --> 00:48:31,210
to the power-- it's I think five
times the number of clauses.

894
00:48:31,210 --> 00:48:36,215
So C is going to be four to
the power of five times number

895
00:48:36,215 --> 00:48:37,590
of clauses because
there are five

896
00:48:37,590 --> 00:48:41,054
of these gadgets per clause.

897
00:48:41,054 --> 00:48:43,220
So at the end we'll just
divide by that and be done.

898
00:48:43,220 --> 00:48:46,503
AUDIENCE: Would it be 10
times because you got it

899
00:48:46,503 --> 00:48:48,848
on the variable side too?

900
00:48:48,848 --> 00:48:50,650
PROFESSOR: Yes 10
times, thank you.

901
00:48:50,650 --> 00:48:52,755
AUDIENCE: Eight-- two
of those don't actually

902
00:48:52,755 --> 00:48:53,630
connect to variables.

903
00:48:53,630 --> 00:48:55,171
PROFESSOR: Two of
them do not connect

904
00:48:55,171 --> 00:48:57,642
to variables, yeah eight.

905
00:48:57,642 --> 00:48:59,333
Eight times the
number of clauses.

906
00:49:02,650 --> 00:49:07,090
All right so now it's
just a matter of-- now

907
00:49:07,090 --> 00:49:10,570
that you understand
what this is now

908
00:49:10,570 --> 00:49:12,980
you could sort of see how
information is communicated.

909
00:49:12,980 --> 00:49:15,940
Because if the variable they
choose is the true setting

910
00:49:15,940 --> 00:49:17,780
it must visit these edges.

911
00:49:17,780 --> 00:49:20,070
And once it touches here
it has to leave here

912
00:49:20,070 --> 00:49:22,430
and this is a edge
going the wrong way.

913
00:49:22,430 --> 00:49:26,480
So you can't try to traverse--
from here you cannot touch

914
00:49:26,480 --> 00:49:28,651
the clauses down below.

915
00:49:28,651 --> 00:49:30,400
Once you touch here
to you have to go here

916
00:49:30,400 --> 00:49:33,370
and then you must
leave here and so on.

917
00:49:33,370 --> 00:49:35,350
But you leave this
one behind and it

918
00:49:35,350 --> 00:49:38,070
must be traversed by the
clause and vice versa.

919
00:49:38,070 --> 00:49:42,090
If I choose this one then these
must be visited by the clauses.

920
00:49:42,090 --> 00:49:44,250
So from the clause
perspective as he

921
00:49:44,250 --> 00:49:47,140
said there are five of these
gadgets, but only three of them

922
00:49:47,140 --> 00:49:47,810
are connected.

923
00:49:47,810 --> 00:49:53,340
So these guys are forced, and
there's a bunch of edges here

924
00:49:53,340 --> 00:49:54,392
and it's a case analysis.

925
00:49:54,392 --> 00:49:55,600
So it'd be the short version.

926
00:49:55,600 --> 00:49:58,350
Let's just do a
couple of examples.

927
00:49:58,350 --> 00:50:00,410
If none of these
have been traversed

928
00:50:00,410 --> 00:50:03,300
by the variable then
the-- pretty much

929
00:50:03,300 --> 00:50:05,820
you have to go straight through.

930
00:50:05,820 --> 00:50:08,030
But then you're in trouble.

931
00:50:08,030 --> 00:50:11,000
Because there's no pointer
back to the beginning.

932
00:50:11,000 --> 00:50:14,840
You can only go back this far.

933
00:50:14,840 --> 00:50:17,590
So if none of-- if you have
to traverse all these things

934
00:50:17,590 --> 00:50:20,350
it's not possible
with the cycle cover.

935
00:50:20,350 --> 00:50:22,740
But if any one of them
has been traversed.

936
00:50:22,740 --> 00:50:25,130
So for example, if this
one has been traversed then

937
00:50:25,130 --> 00:50:28,280
we'll jump over it,
visit these guys,

938
00:50:28,280 --> 00:50:31,430
jump back here, visit this
guy, and jump back there.

939
00:50:31,430 --> 00:50:35,340
And if you check that's the
only way to do it, it's unique.

940
00:50:35,340 --> 00:50:38,190
And similarly any one of
them has been covered,

941
00:50:38,190 --> 00:50:40,690
or if all three of them have
been covered, or if two of them

942
00:50:40,690 --> 00:50:42,148
have been covered
in all cases this

943
00:50:42,148 --> 00:50:44,880
is a unique way from
the clause perspective

944
00:50:44,880 --> 00:50:46,155
to connect all the things.

945
00:50:46,155 --> 00:50:47,530
Now in reality,
there's four ways

946
00:50:47,530 --> 00:50:48,680
to traverse each
of these things.

947
00:50:48,680 --> 00:50:50,346
So the whole thing
grows by this product

948
00:50:50,346 --> 00:50:53,050
but it doesn't matter
which cycle they appear in.

949
00:50:53,050 --> 00:50:56,880
We'll always scale the number
of solutions by factor of four.

950
00:50:56,880 --> 00:51:02,510
So it's nice uniform
scaling C-monious.

951
00:51:02,510 --> 00:51:03,960
And that's permanent.

952
00:51:03,960 --> 00:51:05,410
Pretty cool.

953
00:51:05,410 --> 00:51:07,410
Pretty cool proof.

954
00:51:07,410 --> 00:51:10,340
Now one not so nice
thing about this proof

955
00:51:10,340 --> 00:51:16,640
is it involves numbers negative
one, zero, one, two, and three.

956
00:51:16,640 --> 00:51:18,650
For reasons we will
see in a moment

957
00:51:18,650 --> 00:51:21,260
it'd be really to
just use zero and one.

958
00:51:21,260 --> 00:51:24,520
And it turns out
that's possible,

959
00:51:24,520 --> 00:51:27,240
but it's kind of annoying.

960
00:51:27,240 --> 00:51:28,000
It's nifty though.

961
00:51:28,000 --> 00:51:30,416
I think this will demonstrate
a bunch of fun number theory

962
00:51:30,416 --> 00:51:32,120
things you could do.

963
00:51:32,120 --> 00:51:34,660
So next goal is
zero one permanent.

964
00:51:34,660 --> 00:51:36,830
Now the matrix is
just zero's and one's.

965
00:51:36,830 --> 00:51:41,770
This is sharp P complete.

966
00:51:41,770 --> 00:51:47,508
I'll tell you the reason that
is particularly interesting.

967
00:51:47,508 --> 00:51:49,300
As it gets rid of
the weights it makes

968
00:51:49,300 --> 00:51:57,710
it more clearly-- makes it more
clearly a counting problem.

969
00:51:57,710 --> 00:52:00,440
So first of all is the
number of cycle covers

970
00:52:00,440 --> 00:52:03,300
in the corresponding graph.

971
00:52:03,300 --> 00:52:05,190
No more weights it's
just-- if it's a zero

972
00:52:05,190 --> 00:52:06,100
there's not an edge.

973
00:52:06,100 --> 00:52:07,933
Since you can't traverse
there if it's a one

974
00:52:07,933 --> 00:52:10,120
you can traverse there
and then every cycle cover

975
00:52:10,120 --> 00:52:12,660
will have product one.

976
00:52:12,660 --> 00:52:14,350
So it's just counting them.

977
00:52:14,350 --> 00:52:18,750
But it also happens to be the
number of perfect matchings

978
00:52:18,750 --> 00:52:20,190
in a bipartite graph.

979
00:52:33,380 --> 00:52:36,048
Which bipartite graph?

980
00:52:36,048 --> 00:52:39,480
The bipartite graph where
one side of the bi-partition

981
00:52:39,480 --> 00:52:43,600
is the rows and the other side
is the columns of the matrix.

982
00:52:47,620 --> 00:52:50,290
And you have an edge I j.

983
00:52:55,240 --> 00:53:02,770
If and only if a i sorry
j is not a good not

984
00:53:02,770 --> 00:53:05,270
the best terminology
for I-- where I here's

985
00:53:05,270 --> 00:53:09,660
is a vertex of v one and
j is a vertex of v two.

986
00:53:09,660 --> 00:53:12,390
Whenever I j equals 1, if
it's zero there's no edge.

987
00:53:15,170 --> 00:53:19,530
It's a little confusing because
the matrix is not symmetric

988
00:53:19,530 --> 00:53:23,040
so you might say well, how does
this make an undirected graph?

989
00:53:23,040 --> 00:53:25,890
Because you could have and edge
from i to j, but not to j to I.

990
00:53:25,890 --> 00:53:28,640
That's OK because I is
interpreted in the left space

991
00:53:28,640 --> 00:53:31,010
and j is interpreted
in the right space.

992
00:53:31,010 --> 00:53:33,730
So the asymmetric
case would be the case

993
00:53:33,730 --> 00:53:36,831
that there's an edge here to
here, but not an edge from here

994
00:53:36,831 --> 00:53:37,330
to here.

995
00:53:37,330 --> 00:53:38,850
That's perfectly OK.

996
00:53:38,850 --> 00:53:41,930
This is one, two, three,
for a three by three matrix

997
00:53:41,930 --> 00:53:45,620
we get a six vertex
graph, not three vertex.

998
00:53:45,620 --> 00:53:49,094
That's what allows you to be
asymmetric on a loop, what

999
00:53:49,094 --> 00:53:51,010
we were normally thinking
is a loop just means

1000
00:53:51,010 --> 00:53:53,500
a horizontal edge.

1001
00:53:53,500 --> 00:53:57,000
OK so it turns out if you
look at this graph, which

1002
00:53:57,000 --> 00:53:59,750
is a different graph from
what we started with,

1003
00:53:59,750 --> 00:54:01,950
the number of perfect
matchings in that graph

1004
00:54:01,950 --> 00:54:05,824
is equal to the number of cycle
covers in the other graph,

1005
00:54:05,824 --> 00:54:06,740
in the directed graph.

1006
00:54:06,740 --> 00:54:08,880
So this one's undirected
and bipartite.

1007
00:54:08,880 --> 00:54:12,090
This one was directed
and not bipartite.

1008
00:54:12,090 --> 00:54:15,380
And the rough intuition
is that we're just

1009
00:54:15,380 --> 00:54:17,684
kind of pulling
the vertices apart

1010
00:54:17,684 --> 00:54:20,100
into two versions, the left
version and the right version.

1011
00:54:20,100 --> 00:54:23,010
If you imagine there being
a connection from one

1012
00:54:23,010 --> 00:54:26,740
right to one left and similarly
from two right to two left

1013
00:54:26,740 --> 00:54:28,157
directed.

1014
00:54:28,157 --> 00:54:30,740
And three right to three left,
hey it looks like cycles again.

1015
00:54:30,740 --> 00:54:33,507
You went here, then you went
around, then you went here,

1016
00:54:33,507 --> 00:54:34,340
then you went there.

1017
00:54:34,340 --> 00:54:36,920
That was a cycle and
similarly this is a cycle.

1018
00:54:36,920 --> 00:54:39,390
So they're the same thing.

1019
00:54:39,390 --> 00:54:41,710
This is cool because
perfect matchings

1020
00:54:41,710 --> 00:54:44,150
are interesting in
particular perfect matchings

1021
00:54:44,150 --> 00:54:46,470
are easy to find
in polynomial time.

1022
00:54:46,470 --> 00:54:49,481
But counting them
as Sharp P complete.

1023
00:54:49,481 --> 00:54:50,730
Getting back to that question.

1024
00:54:53,960 --> 00:54:57,990
So that's why we care about zero
one permanence or one reason

1025
00:54:57,990 --> 00:54:58,890
to care about it.

1026
00:54:58,890 --> 00:55:00,312
Let's prove that it's hard.

1027
00:55:16,050 --> 00:55:19,350
And here we'll use a number
theory, pretty basic number

1028
00:55:19,350 --> 00:55:19,850
theory.

1029
00:55:28,600 --> 00:55:30,320
So first claim is
that computing the

1030
00:55:30,320 --> 00:55:35,600
permanent of a matrix, general
matrix nonzero one modulo

1031
00:55:35,600 --> 00:55:37,840
given integer r is hard.

1032
00:55:42,260 --> 00:55:44,200
r has to be an input
for this problem.

1033
00:55:44,200 --> 00:55:47,055
It's not like computing at
Mod three necessarily as hard,

1034
00:55:47,055 --> 00:55:50,977
but computing at modular
anything is hard.

1035
00:55:50,977 --> 00:55:53,560
And here we're going to finally
use some multi-color reduction

1036
00:55:53,560 --> 00:55:54,910
power.

1037
00:55:54,910 --> 00:55:58,340
So the idea is, suppose you
could solve this problem then

1038
00:55:58,340 --> 00:56:00,551
I claim I can solve permanent.

1039
00:56:00,551 --> 00:56:01,700
Anyone know how?

1040
00:56:01,700 --> 00:56:03,200
AUDIENCE: Chinese
remainder theorem?

1041
00:56:03,200 --> 00:56:15,927
PROFESSOR: Chinese
remainder theorem

1042
00:56:15,927 --> 00:56:18,010
OK if you don't know the
Chinese remainder theorem

1043
00:56:18,010 --> 00:56:19,030
but you should know it.

1044
00:56:19,030 --> 00:56:21,490
It's good.

1045
00:56:21,490 --> 00:56:24,740
So the idea is
we're going to set r

1046
00:56:24,740 --> 00:56:33,970
to be all primes up to how big?

1047
00:56:33,970 --> 00:56:38,230
Or we can stop when the
product of all the primes

1048
00:56:38,230 --> 00:56:44,060
that we've considered is bigger
than the potential permanent.

1049
00:56:44,060 --> 00:56:49,190
And the permanent is at
most m to the n times

1050
00:56:49,190 --> 00:56:52,350
n factorial, where
m is the largest

1051
00:56:52,350 --> 00:56:53,640
absolute value in the matrix.

1052
00:57:00,350 --> 00:57:03,480
That's one upper bound,
doesn't really matter.

1053
00:57:03,480 --> 00:57:05,110
But the point is,
it is computable

1054
00:57:05,110 --> 00:57:07,560
and if you take the
logarithm it's not so big.

1055
00:57:07,560 --> 00:57:09,490
All these numbers
are at least two.

1056
00:57:09,490 --> 00:57:11,550
All primes are at least two.

1057
00:57:11,550 --> 00:57:15,900
So the number of primes
we'll have to consider

1058
00:57:15,900 --> 00:57:19,480
is at most log base 2 of that.

1059
00:57:19,480 --> 00:57:25,150
So this means the
number of primes,

1060
00:57:25,150 --> 00:57:29,000
so that's log of m to
the n times n factorial,

1061
00:57:29,000 --> 00:57:32,700
and this is roughly--
I'll put a total here.

1062
00:57:32,700 --> 00:57:36,400
This is like m
log n that's exact

1063
00:57:36,400 --> 00:57:42,200
plus n log n that's approximate
but it's minus order n

1064
00:57:42,200 --> 00:57:44,530
so it won't hurt us.

1065
00:57:44,530 --> 00:57:46,400
So that's good,
that's a polynomial.

1066
00:57:46,400 --> 00:57:50,040
Log m is a reasonable thing,
m was given us as a number

1067
00:57:50,040 --> 00:57:52,680
so log n is part of this input.

1068
00:57:52,680 --> 00:57:57,860
So as our input and n is
obviously good as the dimension

1069
00:57:57,860 --> 00:57:58,640
of the matrix.

1070
00:57:58,640 --> 00:58:01,274
So this is a polynomial
number of calls.

1071
00:58:01,274 --> 00:58:02,940
We're going to compute
the permanent mod

1072
00:58:02,940 --> 00:58:05,580
r for all these things and
our Chinese remainder theorem

1073
00:58:05,580 --> 00:58:08,850
tells you if you
know a number which

1074
00:58:08,850 --> 00:58:12,920
is the permanent module
all primes whose products--

1075
00:58:12,920 --> 00:58:16,770
Well if you know a number of
modulo a bunch of primes then

1076
00:58:16,770 --> 00:58:21,550
you can figure out the number of
modulo, the product of primes.

1077
00:58:21,550 --> 00:58:24,290
And if we set the product to
be bigger than the number could

1078
00:58:24,290 --> 00:58:27,730
be, then we know [? any ?]
modulo that is actually

1079
00:58:27,730 --> 00:58:29,500
knowing the number itself.

1080
00:58:29,500 --> 00:58:31,820
So then we can
reconstruct the permanent.

1081
00:58:35,540 --> 00:58:37,420
So here we're really
using multicall.

1082
00:58:45,984 --> 00:58:47,400
I think that's the
first time I've

1083
00:58:47,400 --> 00:58:50,150
used Chinese remainder theorem
in a class that I've taught.

1084
00:58:53,101 --> 00:58:53,600
Good stuff.

1085
00:58:57,837 --> 00:59:00,295
AUDIENCE: Do you know why it's
called the Chinese remainder

1086
00:59:00,295 --> 00:59:00,860
theorem?

1087
00:59:00,860 --> 00:59:02,830
PROFESSOR: I assume
it was proved

1088
00:59:02,830 --> 00:59:06,040
by a Chinese mathematician
but I-- anyone know?

1089
00:59:06,040 --> 00:59:07,291
AUDIENCE: Sun Tzu.

1090
00:59:07,291 --> 00:59:08,460
PROFESSOR: Sun Tzu.

1091
00:59:08,460 --> 00:59:12,645
OK I've heard of him.

1092
00:59:12,645 --> 00:59:14,650
AUDIENCE: Not the guy
who did Art of War.

1093
00:59:14,650 --> 00:59:15,400
PROFESSOR: Oh, OK.

1094
00:59:15,400 --> 00:59:17,800
I don't know him then.

1095
00:59:17,800 --> 00:59:21,100
Another Sun Tzu.

1096
00:59:21,100 --> 00:59:23,970
Cool, yeah so permanent
mod r is hard.

1097
00:59:23,970 --> 00:59:32,400
Next claim is that the zero
one permanent mod r is hard.

1098
00:59:34,932 --> 00:59:36,390
The whole reason
we're going to mod

1099
00:59:36,390 --> 00:59:38,473
r-- I mean it's interesting
to know that computing

1100
00:59:38,473 --> 00:59:40,680
permanent mod r is just
as hard as permanent

1101
00:59:40,680 --> 00:59:43,710
but it's kind of standard thing.

1102
00:59:43,710 --> 00:59:45,610
The reason I wanted
to go to mod r

1103
00:59:45,610 --> 00:59:47,800
was to get rid of
these negative numbers.

1104
00:59:47,800 --> 00:59:49,730
Negative numbers are annoying.

1105
00:59:49,730 --> 00:59:54,960
Once I go to mod any r
negative numbers flip around

1106
00:59:54,960 --> 00:59:58,480
and I can figure everything
is positive or non-negative.

1107
00:59:58,480 --> 01:00:01,680
Now I can go back
to gadget land.

1108
01:00:01,680 --> 01:00:04,220
So suppose I have an
edge of weight five,

1109
01:00:04,220 --> 01:00:06,270
this will work for any
number greater than one.

1110
01:00:06,270 --> 01:00:08,380
If It's one I don't touch it.

1111
01:00:08,380 --> 01:00:11,680
If it's greater than one I'm
going to make this thing.

1112
01:00:14,900 --> 01:00:19,360
And the claim is there
are exactly five ways

1113
01:00:19,360 --> 01:00:20,390
to use this edge.

1114
01:00:20,390 --> 01:00:22,470
Either you don't use it
and you don't use it.

1115
01:00:22,470 --> 01:00:26,290
But if you do use it,
exactly five ways to use it.

1116
01:00:26,290 --> 01:00:29,229
If you make this traversal,
there are five ways to do it.

1117
01:00:29,229 --> 01:00:30,770
Remember this has
to be a cycle cover

1118
01:00:30,770 --> 01:00:32,860
so we have to visit everything.

1119
01:00:32,860 --> 01:00:36,450
If you come up here, can't go
that way gotta go straight.

1120
01:00:36,450 --> 01:00:38,295
Can't go that way,
got to go straight.

1121
01:00:38,295 --> 01:00:40,580
Hmm, OK.

1122
01:00:40,580 --> 01:00:42,830
So in fact, the rest
of this cycle cover

1123
01:00:42,830 --> 01:00:44,910
must be entirely
within this picture.

1124
01:00:44,910 --> 01:00:47,210
And so for example, you
could say that's a cycle

1125
01:00:47,210 --> 01:00:49,830
and then that's a
cycle, these are loops.

1126
01:00:49,830 --> 01:00:51,950
And then that's a
cycle and then that's

1127
01:00:51,950 --> 01:00:55,390
a cycle and then that's a
cycle and then that's a cycle.

1128
01:00:55,390 --> 01:00:57,620
And whoops I'm left
with one vertex.

1129
01:00:57,620 --> 01:01:00,930
So in fact I have to choose
exactly one of the loops,

1130
01:01:00,930 --> 01:01:03,160
put that in and the
rest can be covered.

1131
01:01:03,160 --> 01:01:06,202
For example, if I choose this
guy then this will be a cycle,

1132
01:01:06,202 --> 01:01:07,910
this'll be a cycle,
this will be a cycle.

1133
01:01:07,910 --> 01:01:10,490
Perfect parity this will be
a cycle, is will be a cycle,

1134
01:01:10,490 --> 01:01:11,750
and this will be a cycle.

1135
01:01:11,750 --> 01:01:12,730
Can't choose two loops.

1136
01:01:12,730 --> 01:01:14,271
If I chose like
these two groups then

1137
01:01:14,271 --> 01:01:15,960
this guy would be isolated.

1138
01:01:15,960 --> 01:01:18,140
So you have to choose
exactly one loop

1139
01:01:18,140 --> 01:01:19,710
and there are
exactly five of them.

1140
01:01:19,710 --> 01:01:23,890
In general you make-- there'd be
k of them if you want weight k,

1141
01:01:23,890 --> 01:01:25,550
and you can simulate.

1142
01:01:25,550 --> 01:01:29,030
If you don't use this edge then
there's actually only one way

1143
01:01:29,030 --> 01:01:29,610
to do it.

1144
01:01:29,610 --> 01:01:33,050
You have to go all the
way around like this.

1145
01:01:33,050 --> 01:01:33,762
So that's cool.

1146
01:01:33,762 --> 01:01:35,970
If you don't use the edge,
didn't mess with anything.

1147
01:01:35,970 --> 01:01:38,097
If you use the edge you
get a weight of five.

1148
01:01:38,097 --> 01:01:40,180
Multiplicative, because
this is independent of all

1149
01:01:40,180 --> 01:01:42,170
the other such choices.

1150
01:01:42,170 --> 01:01:47,490
So you can simulate a
high-weight edge, module r.

1151
01:01:47,490 --> 01:01:49,990
In general you can simulate
a non-negative weight edge

1152
01:01:49,990 --> 01:01:51,460
like this.

1153
01:01:51,460 --> 01:01:54,030
Just these are all weight one
obviously and absent edges

1154
01:01:54,030 --> 01:01:55,310
are zero.

1155
01:01:55,310 --> 01:01:57,940
So we can convert
in the module r

1156
01:01:57,940 --> 01:02:01,840
setting there are no
negative numbers essentially.

1157
01:02:01,840 --> 01:02:03,885
So we can just do that
simulation, everything

1158
01:02:03,885 --> 01:02:05,820
will work mod r.

1159
01:02:05,820 --> 01:02:07,110
So use gadget.

1160
01:02:16,530 --> 01:02:19,300
Now finally we can prove
zero one permanent is hard.

1161
01:02:24,500 --> 01:02:25,010
Why?

1162
01:02:25,010 --> 01:02:27,093
Because if we could compute
the zero one permanent

1163
01:02:27,093 --> 01:02:28,730
we can compute at mod r.

1164
01:02:28,730 --> 01:02:32,290
Just compute the permanent,
take the answer mod r and you

1165
01:02:32,290 --> 01:02:34,250
solve zero one permanent mod r.

1166
01:02:34,250 --> 01:02:37,410
So this is what I might
call a one-call reduction.

1167
01:02:37,410 --> 01:02:39,550
It's not our usual notion
of one-call reduction

1168
01:02:39,550 --> 01:02:42,990
because we're doing
stuff at the end.

1169
01:02:42,990 --> 01:02:45,520
But I'm going to use
one-call in this setting.

1170
01:02:50,700 --> 01:02:53,630
Just call it-- in fact we don't
even have to change the input.

1171
01:02:53,630 --> 01:02:55,600
Then we get the output
when we computed mod r.

1172
01:02:55,600 --> 01:02:56,100
Question.

1173
01:02:56,100 --> 01:02:59,920
AUDIENCE: So in
that the weight you

1174
01:02:59,920 --> 01:03:01,348
can have really high weight.

1175
01:03:01,348 --> 01:03:04,522
But then your
[INAUDIBLE] nodes so then

1176
01:03:04,522 --> 01:03:08,387
when you use that wouldn't n?

1177
01:03:08,387 --> 01:03:09,220
PROFESSOR: Oh I see.

1178
01:03:09,220 --> 01:03:10,845
If the weights were
exponentially large

1179
01:03:10,845 --> 01:03:12,350
this would be bad news.

1180
01:03:12,350 --> 01:03:13,940
But they're not
exponentially large

1181
01:03:13,940 --> 01:03:17,000
because we know they're
all at most three.

1182
01:03:17,000 --> 01:03:20,570
They're between
negative one and three.

1183
01:03:20,570 --> 01:03:26,760
Oh but negative
one-- Yes, OK good

1184
01:03:26,760 --> 01:03:28,300
I need the prime number theorem.

1185
01:03:28,300 --> 01:03:28,800
Thank you.

1186
01:03:28,800 --> 01:03:32,220
So we have numbers
negative one then

1187
01:03:32,220 --> 01:03:34,150
we're mapping them
modulo some prime.

1188
01:03:34,150 --> 01:03:38,090
Now negative one is actually
the prime minus one.

1189
01:03:38,090 --> 01:03:43,440
So indeed, we do get weights
that are as large as r.

1190
01:03:43,440 --> 01:03:47,870
So this is-- r here we're
assuming is encoded in unary.

1191
01:03:51,610 --> 01:03:55,710
It turns out we can afford
to encode the primes in unary

1192
01:03:55,710 --> 01:03:59,130
because the number
of primes that we use

1193
01:03:59,130 --> 01:04:04,370
is this polynomial thing,
a weakly polynomial thing.

1194
01:04:04,370 --> 01:04:09,670
And by the prime number theorem,
the prime with that index

1195
01:04:09,670 --> 01:04:11,000
is roughly that big.

1196
01:04:11,000 --> 01:04:12,950
It's the log factor larger.

1197
01:04:12,950 --> 01:04:16,550
So the primes are going
to-- the actual value

1198
01:04:16,550 --> 01:04:21,860
of the prime is going to
be something like n log

1199
01:04:21,860 --> 01:04:32,270
m, times log base e, and log m.

1200
01:04:32,270 --> 01:04:35,470
By the prime number theorem and
that's again weakly polynomial.

1201
01:04:35,470 --> 01:04:37,520
And so we can assume
ours encoded in unary.

1202
01:04:41,450 --> 01:04:45,643
Wow, we get to use the prime
number theorem that's fun.

1203
01:04:45,643 --> 01:04:48,226
AUDIENCE: Is that the first time
you had to use prime numbers?

1204
01:04:48,226 --> 01:04:50,511
PROFESSOR: Probably.

1205
01:04:50,511 --> 01:04:52,760
Pretty sure I've used prime
number theorem if and only

1206
01:04:52,760 --> 01:04:55,630
if I've used Chinese
remainder theorem.

1207
01:04:55,630 --> 01:04:56,620
They go hand in hand.

1208
01:04:59,430 --> 01:04:59,930
Clear?

1209
01:04:59,930 --> 01:05:02,564
So this was kind
of roundabout way,

1210
01:05:02,564 --> 01:05:04,730
but we ended up getting rid
of the negative numbers.

1211
01:05:04,730 --> 01:05:06,230
Luckily it still worked.

1212
01:05:06,230 --> 01:05:08,210
And now it's zero
one permanence hard,

1213
01:05:08,210 --> 01:05:10,890
therefore counting the number
of perfect matchings in a given

1214
01:05:10,890 --> 01:05:12,890
bipartite graph isn't
going to be hard.

1215
01:05:12,890 --> 01:05:15,464
These problems were equivalent,
like they're identical.

1216
01:05:15,464 --> 01:05:17,380
So you could-- this was
a reduction in one way

1217
01:05:17,380 --> 01:05:20,980
but that's equally
reducible both ways.

1218
01:05:20,980 --> 01:05:23,430
So you can reduce this
one, reduce this one

1219
01:05:23,430 --> 01:05:27,306
to perfect matchings
or vice versa.

1220
01:05:27,306 --> 01:05:27,805
All right.

1221
01:05:31,047 --> 01:05:32,630
Here's some more fun
things we can do.

1222
01:05:50,000 --> 01:05:51,480
Guess I can add to my list here.

1223
01:05:51,480 --> 01:05:55,600
But in particular we
have zero one permanent.

1224
01:06:00,200 --> 01:06:04,141
So there are other ways you
might be interested in counting

1225
01:06:04,141 --> 01:06:04,640
matchings.

1226
01:06:12,010 --> 01:06:15,380
So, so far we know it's hard
to count perfect matchings.

1227
01:06:15,380 --> 01:06:20,774
So in a balanced bipartite graph
we had n vertices on the left,

1228
01:06:20,774 --> 01:06:21,940
and n vertices on the right.

1229
01:06:21,940 --> 01:06:23,790
We're going to use that.

1230
01:06:23,790 --> 01:06:26,790
Then the hope is that
there's matching a size n.

1231
01:06:26,790 --> 01:06:29,129
But in general, you could
just count maybe those

1232
01:06:29,129 --> 01:06:29,920
don't exist at all.

1233
01:06:29,920 --> 01:06:32,060
Maybe we just want to
count maximal matchings.

1234
01:06:32,060 --> 01:06:34,280
These are not maximum matchings.

1235
01:06:34,280 --> 01:06:37,340
Maximal, meaning you can't
add any more edges to them.

1236
01:06:37,340 --> 01:06:40,800
They're locally maximum.

1237
01:06:40,800 --> 01:06:43,140
So that's going to
be a bigger number,

1238
01:06:43,140 --> 01:06:45,880
it's going to be always
bigger than zero.

1239
01:06:45,880 --> 01:06:47,850
Because the empty
matching, well you

1240
01:06:47,850 --> 01:06:49,308
could start with
the empty matching

1241
01:06:49,308 --> 01:06:52,000
and add things you'll get at
least one maximal matching.

1242
01:06:52,000 --> 01:06:54,150
But this is also
sharp P complete,

1243
01:06:54,150 --> 01:06:56,090
and you can prove
it using some tricks

1244
01:06:56,090 --> 01:07:01,380
from bipartite perfect
matching, counting.

1245
01:07:04,580 --> 01:07:06,070
Don't have a ton
of time so I'll go

1246
01:07:06,070 --> 01:07:08,840
through this relatively quick.

1247
01:07:08,840 --> 01:07:12,380
We're going to take each vertex
and convert it, basically

1248
01:07:12,380 --> 01:07:14,920
make n copies of it.

1249
01:07:14,920 --> 01:07:18,760
And when we have an edge
that's going to turn

1250
01:07:18,760 --> 01:07:23,940
into a biclique between them.

1251
01:07:23,940 --> 01:07:25,800
Why did I address
such a big one?

1252
01:07:29,970 --> 01:07:32,650
This was supposed to be n and n.

1253
01:07:32,650 --> 01:07:34,220
OK so just blow up every edge.

1254
01:07:34,220 --> 01:07:37,850
My intent is to make matchings
become more plentiful.

1255
01:07:40,480 --> 01:07:47,552
So in general, if I used to
have a matching of size i,

1256
01:07:47,552 --> 01:07:54,310
I end up converting it into n
factorial to the i-th power,

1257
01:07:54,310 --> 01:08:04,460
distinct matchings
of size n times i.

1258
01:08:04,460 --> 01:08:07,480
OK because if I
use an edge here,

1259
01:08:07,480 --> 01:08:09,790
now I get to put in a
arbitrary perfect matching

1260
01:08:09,790 --> 01:08:14,160
in this biclique and
they're n factorial of them.

1261
01:08:14,160 --> 01:08:17,229
Cool, why did I do that?

1262
01:08:17,229 --> 01:08:19,410
Because now I
suppose that I knew

1263
01:08:19,410 --> 01:08:22,627
how to count the number
of maximum actions.

1264
01:08:22,627 --> 01:08:24,710
So they're going to be
matchings of various sizes.

1265
01:08:24,710 --> 01:08:27,740
From that I want to extract the
number of perfect matchings.

1266
01:08:27,740 --> 01:08:31,170
Sounds impossible, but
when you make things giant

1267
01:08:31,170 --> 01:08:33,785
like this they kind of-- all
the matchings kind of separate.

1268
01:08:33,785 --> 01:08:37,029
It's like biology or something.

1269
01:08:37,029 --> 01:08:37,560
Chemistry.

1270
01:08:37,560 --> 01:08:43,260
So let's see, it
shows how much I know.

1271
01:08:43,260 --> 01:08:49,790
Number of maximal
matchings is going

1272
01:08:49,790 --> 01:08:55,420
to be sum i equals
zero to n over 2.

1273
01:08:55,420 --> 01:08:58,800
That's the possible
sizes of the matchings.

1274
01:08:58,800 --> 01:09:00,370
Of the old matchings
I should say.

1275
01:09:00,370 --> 01:09:05,670
Number of original
maximal matchings

1276
01:09:05,670 --> 01:09:15,290
in the input graph of size i,
times n factorial to the i.

1277
01:09:15,290 --> 01:09:16,830
OK this is just rewriting.

1278
01:09:16,830 --> 01:09:19,189
I'm just this thing, but
I'm summing over all i.

1279
01:09:19,189 --> 01:09:21,939
So I have however many matchings
I used to have a size i,

1280
01:09:21,939 --> 01:09:24,010
but then I multiply them
by n factorial to the i,

1281
01:09:24,010 --> 01:09:25,426
because these are
all independent.

1282
01:09:29,970 --> 01:09:33,319
And this thing,
the original number

1283
01:09:33,319 --> 01:09:36,770
of matchings in the worst case,
I mean the largest it could be

1284
01:09:36,770 --> 01:09:38,529
is when everything
is a biclique.

1285
01:09:38,529 --> 01:09:47,275
And then we have n over two
factorial maximal matchings

1286
01:09:47,275 --> 01:09:47,775
originally.

1287
01:09:50,569 --> 01:09:54,650
And this is smaller than that.

1288
01:09:54,650 --> 01:09:58,920
And therefore we can pull
this apart as a number modulo

1289
01:09:58,920 --> 01:10:00,650
n factorial.

1290
01:10:00,650 --> 01:10:05,020
And each digit is the number of
maximum actions of each size.

1291
01:10:05,020 --> 01:10:07,630
And we look at
the last digit, we

1292
01:10:07,630 --> 01:10:11,800
get all the number of
maximum matchings of size n

1293
01:10:11,800 --> 01:10:16,140
over two also known as the
number of perfect matchings.

1294
01:10:16,140 --> 01:10:16,640
Question?

1295
01:10:16,640 --> 01:10:18,970
AUDIENCE: So that
second max is maximal?

1296
01:10:18,970 --> 01:10:20,340
PROFESSOR: Yes this is maximal.

1297
01:10:24,680 --> 01:10:26,360
Other questions?

1298
01:10:26,360 --> 01:10:29,000
So again I kind of numbered
[? through ?] trick

1299
01:10:29,000 --> 01:10:33,270
by blowing up-- well
by making each of these

1300
01:10:33,270 --> 01:10:36,130
get multiplied by a
different huge number

1301
01:10:36,130 --> 01:10:39,070
we can pull them apart,
separate them-- In logarithm

1302
01:10:39,070 --> 01:10:42,595
these numbers are not huge
so it is actually plausible

1303
01:10:42,595 --> 01:10:43,850
you could compute this thing.

1304
01:10:46,699 --> 01:10:48,490
And a tree graph you
definitely could do it

1305
01:10:48,490 --> 01:10:51,950
by various multiplications
at each level.

1306
01:10:51,950 --> 01:10:53,300
All right.

1307
01:10:53,300 --> 01:10:58,475
So that's nice, let's do
another one of that flavor.

1308
01:11:01,163 --> 01:11:02,704
AUDIENCE: So I was
actually wondering

1309
01:11:02,704 --> 01:11:04,745
can you find all those
primes in polynomial time?

1310
01:11:04,745 --> 01:11:07,820
PROFESSOR: You could use
the Sieve of Eratosthenes.

1311
01:11:07,820 --> 01:11:09,836
AUDIENCE: Is that
like going to be--

1312
01:11:09,836 --> 01:11:12,460
PROFESSOR: We can handle pseudo
poly here because as we argued,

1313
01:11:12,460 --> 01:11:14,230
the primes are not that big.

1314
01:11:14,230 --> 01:11:16,650
So we could just
spend-- we could just

1315
01:11:16,650 --> 01:11:18,650
do Sieve of Eratosthenes,
we can afford

1316
01:11:18,650 --> 01:11:21,620
to spend a linear time
on the largest prime.

1317
01:11:21,620 --> 01:11:22,725
Or quadratic in that even.

1318
01:11:22,725 --> 01:11:24,100
You could do the
naive algorithm.

1319
01:11:29,510 --> 01:11:33,280
We're not doing real Hogan's
[? mid-number ?] theory here.

1320
01:11:33,280 --> 01:11:37,250
I could tell where you
have to be more careful.

1321
01:11:37,250 --> 01:11:40,190
All right. .

1322
01:11:40,190 --> 01:11:43,492
So here's another question.

1323
01:11:43,492 --> 01:11:45,700
What if I just want to count
the number of matchings?

1324
01:11:45,700 --> 01:11:46,691
No condition?

1325
01:11:50,930 --> 01:11:54,990
This is going to use
almost the reverse trick.

1326
01:11:54,990 --> 01:11:57,800
So no maximal constraint
just all matchings,

1327
01:11:57,800 --> 01:11:59,050
including the empty matchings.

1328
01:11:59,050 --> 01:12:01,100
Count them all.

1329
01:12:01,100 --> 01:12:04,970
We'll see why in a moment,
this is quite natural.

1330
01:12:04,970 --> 01:12:09,660
So I claim I can do a multicall
reduction to bipartite

1331
01:12:09,660 --> 01:12:11,780
number of maximal matchings.

1332
01:12:11,780 --> 01:12:15,110
And here are my calls
for every graph g,

1333
01:12:15,110 --> 01:12:18,190
I'm going to make
a graph g prime.

1334
01:12:18,190 --> 01:12:21,020
Where if I have a
vertex I'm going to add

1335
01:12:21,020 --> 01:12:25,130
k extra nodes connected
just to that vertex.

1336
01:12:25,130 --> 01:12:28,022
So these are leaves
and so if this vertex

1337
01:12:28,022 --> 01:12:30,230
was unmatched over here,
then I have k different ways

1338
01:12:30,230 --> 01:12:32,580
to match it over here.

1339
01:12:32,580 --> 01:12:36,840
And my intent is to measure this
number of matchings of size n

1340
01:12:36,840 --> 01:12:39,080
over two minus k.

1341
01:12:39,080 --> 01:12:40,060
That would be the hope.

1342
01:12:44,500 --> 01:12:53,110
So if I have m r matchings
of size r over here,

1343
01:12:53,110 --> 01:12:58,820
these will get mapped
to Mr times k plus one

1344
01:12:58,820 --> 01:13:09,110
to the r, matchings of
size no, not of size.

1345
01:13:09,110 --> 01:13:10,040
Matchings over here.

1346
01:13:16,654 --> 01:13:18,820
Because for each one I can
either leave it unmatched

1347
01:13:18,820 --> 01:13:20,620
or I could add this edge,
or I could add this edge,

1348
01:13:20,620 --> 01:13:22,078
or I could add this
edge and that's

1349
01:13:22,078 --> 01:13:24,060
true for every unmatched vertex.

1350
01:13:24,060 --> 01:13:28,062
Sorry, this is not size r, this
is size n over two minus r.

1351
01:13:28,062 --> 01:13:30,660
R is going to be the
number of leftover guys.

1352
01:13:30,660 --> 01:13:33,610
Then they kind of
pops out over here.

1353
01:13:33,610 --> 01:13:36,010
So I'm going to run
this algorithm-- I'm

1354
01:13:36,010 --> 01:13:43,040
going to compute the number of
matchings in Gk for all k up

1355
01:13:43,040 --> 01:13:49,250
to like n over two plus one.

1356
01:13:49,250 --> 01:14:00,930
So I'm going to do
this and what I end up

1357
01:14:00,930 --> 01:14:07,390
computing is number of matchings
in each Gk, which is going

1358
01:14:07,390 --> 01:14:10,022
to be the sum over all r.

1359
01:14:10,022 --> 01:14:15,885
r is zero to n over two
of Mr the original number

1360
01:14:15,885 --> 01:14:19,830
of matchings the size
n over two minus r,

1361
01:14:19,830 --> 01:14:22,510
times k plus one to the r.

1362
01:14:22,510 --> 01:14:28,170
Now this is a polynomial
in k plus one.

1363
01:14:28,170 --> 01:14:33,000
And if I evaluate a polynomial
at its degree plus one

1364
01:14:33,000 --> 01:14:35,500
different points I can
reconstruct the coefficients

1365
01:14:35,500 --> 01:14:36,720
of the polynomial.

1366
01:14:36,720 --> 01:14:39,160
And therefore get all of
these and in particular

1367
01:14:39,160 --> 01:14:44,980
get m zero, which is the
number of perfect matchings.

1368
01:14:44,980 --> 01:14:45,980
Ta da.

1369
01:14:58,130 --> 01:14:58,990
Having too much fun.

1370
01:15:06,450 --> 01:15:07,190
Enough matchings.

1371
01:15:10,700 --> 01:15:13,720
We have all these versions
of SAT, which are hard.

1372
01:15:13,720 --> 01:15:17,230
But in fact there are
even funnier versions

1373
01:15:17,230 --> 01:15:19,360
like positive 2SAT.

1374
01:15:22,850 --> 01:15:25,600
Positive 2SAT is
really easy to decide,

1375
01:15:25,600 --> 01:15:29,510
but it turns out if I add a
sharp it is sharp P complete.

1376
01:15:29,510 --> 01:15:31,402
This is sort of like
max 2SAT, but here we

1377
01:15:31,402 --> 01:15:32,860
have to satisfy
all the constraints

1378
01:15:32,860 --> 01:15:34,985
but you want to count how
many different ways there

1379
01:15:34,985 --> 01:15:37,090
are to solve it.

1380
01:15:37,090 --> 01:15:39,350
This is the same thing
as vertex cover remember.

1381
01:15:39,350 --> 01:15:41,550
Vertex cover's the
same as positive 2SAT.

1382
01:15:41,550 --> 01:15:43,730
So also sharp
vertex cover's hard

1383
01:15:43,730 --> 01:15:46,570
and therefore also
sharp clique is hard.

1384
01:15:46,570 --> 01:15:49,300
So this is actually a
parsimonious reduction

1385
01:15:49,300 --> 01:15:51,275
from bipartite matching.

1386
01:15:51,275 --> 01:15:53,390
That's why I wanted
to get there.

1387
01:15:53,390 --> 01:16:00,710
So for every edge we're going
to make a variable, x of e,

1388
01:16:00,710 --> 01:16:05,500
which is true if the edge is
not in the perfect matching.

1389
01:16:05,500 --> 01:16:11,230
And so then whenever I have
two incident edges in f,

1390
01:16:11,230 --> 01:16:14,440
we're going to have a clause
which is either I don't use e,

1391
01:16:14,440 --> 01:16:16,610
or I don't use f.

1392
01:16:16,610 --> 01:16:18,424
That's 2SAT, done.

1393
01:16:23,570 --> 01:16:24,220
Satisfying?

1394
01:16:24,220 --> 01:16:25,050
It's parsimonious.

1395
01:16:25,050 --> 01:16:27,330
So satisfying assignments
will be one to one

1396
01:16:27,330 --> 01:16:29,730
with bipartite matchings.

1397
01:16:29,730 --> 01:16:31,660
So this is why you should care.

1398
01:16:31,660 --> 01:16:35,600
And if we instead reduce
from-- did I erase it?

1399
01:16:35,600 --> 01:16:40,450
Bipartite maximal
matchings up here,

1400
01:16:40,450 --> 01:16:43,220
then I get the
counting the number

1401
01:16:43,220 --> 01:16:51,530
of maximally true assignments
that satisfy the 2SAT clause

1402
01:16:51,530 --> 01:16:54,430
is also hard.

1403
01:16:54,430 --> 01:16:58,120
For what it's worth, a
different way of counting that.

1404
01:16:58,120 --> 01:16:58,820
(BREAK IN VIDEO)

1405
01:16:58,820 --> 01:17:01,710
Maxwell means you can't set
any more variables true.

1406
01:17:01,710 --> 01:17:02,210
Yeah.

1407
01:17:02,210 --> 01:17:05,978
AUDIENCE: Since the edges on
your positive 2SAT reduction

1408
01:17:05,978 --> 01:17:13,154
are the variables there are
true when the edge is not

1409
01:17:13,154 --> 01:17:14,122
in a matching.

1410
01:17:14,122 --> 01:17:15,962
I think it's maximally false.

1411
01:17:15,962 --> 01:17:16,920
And not maximally true.

1412
01:17:16,920 --> 01:17:19,740
Also maximally true, you just
said everything was true.

1413
01:17:19,740 --> 01:17:20,810
PROFESSOR: Yes.

1414
01:17:20,810 --> 01:17:22,390
Right, right, right, sorry.

1415
01:17:22,390 --> 01:17:23,223
I see, you're right.

1416
01:17:23,223 --> 01:17:25,450
So it's minimal solutions
for a 2SAT, thank you.

1417
01:17:28,700 --> 01:17:40,650
OK in fact, it's known that
three regular bipartite planar

1418
01:17:40,650 --> 01:17:41,720
sharp vertex cover.

1419
01:17:46,262 --> 01:17:47,220
I won't prove this one.

1420
01:17:47,220 --> 01:17:50,230
But in particular you
can make this planar,

1421
01:17:50,230 --> 01:17:52,420
although it doesn't look
like we have positive here.

1422
01:17:52,420 --> 01:17:54,461
And you can also make it
bipartite, which doesn't

1423
01:17:54,461 --> 01:17:55,810
mean a lot in 2SAT land.

1424
01:17:55,810 --> 01:17:59,830
But it means-- makes a lot of
sense in vertex cover land.

1425
01:17:59,830 --> 01:18:01,620
In my zero remaining
minutes I want

1426
01:18:01,620 --> 01:18:03,030
to mention one more concept.

1427
01:18:05,736 --> 01:18:09,560
To go back to the
original question

1428
01:18:09,560 --> 01:18:13,400
of uniqueness for puzzles.

1429
01:18:13,400 --> 01:18:16,290
And you'll see this in the
literature if you look at it.

1430
01:18:16,290 --> 01:18:18,730
Lot of people won't
even mention sharp P,

1431
01:18:18,730 --> 01:18:21,690
but they'll mention ASP.

1432
01:18:21,690 --> 01:18:25,880
ASP is slightly weaker but
for most intents and purposes

1433
01:18:25,880 --> 01:18:28,210
essentially identical
notion to sharp P,

1434
01:18:28,210 --> 01:18:30,300
from a hardness perspective.

1435
01:18:30,300 --> 01:18:33,270
The goal for-- in general,
if I have a problem

1436
01:18:33,270 --> 01:18:35,500
a search problem
like we started with.

1437
01:18:35,500 --> 01:18:38,650
The ASP version of
that search problem is,

1438
01:18:38,650 --> 01:18:43,180
I give you a solution and I want
to know is there another one?

1439
01:18:43,180 --> 01:18:44,930
This is a little
different from everything

1440
01:18:44,930 --> 01:18:48,085
we've seen because I actually
give you a starting point.

1441
01:18:48,085 --> 01:18:49,960
There's some problems
where this helps a lot.

1442
01:18:49,960 --> 01:18:54,110
For example, if the solution
is never unique like coloring,

1443
01:18:54,110 --> 01:18:55,890
I give you a k
coloring, I can give you

1444
01:18:55,890 --> 01:18:58,310
another one I'll just
swap colors one and two.

1445
01:18:58,310 --> 01:19:01,890
Or more subtle if I give you
a Hamiltonian cycle in a three

1446
01:19:01,890 --> 01:19:02,610
regular graph.

1447
01:19:02,610 --> 01:19:06,860
There's always a way to switch
it and make another one.

1448
01:19:06,860 --> 01:19:09,940
So ASP is sometimes easy.

1449
01:19:09,940 --> 01:19:13,630
But you would basically
use the same reductions.

1450
01:19:13,630 --> 01:19:16,700
If you can find a
parsimonious reduction that

1451
01:19:16,700 --> 01:19:19,310
also has the property--
so parsimonious reduction

1452
01:19:19,310 --> 01:19:22,490
there is a one to one
bijection between solutions

1453
01:19:22,490 --> 01:19:23,900
to x and solutions to x prime.

1454
01:19:23,900 --> 01:19:27,382
If you could also compute
that bijection, if you can

1455
01:19:27,382 --> 01:19:29,090
convert a solution
from one to the other.

1456
01:19:29,090 --> 01:19:31,080
Which we could do in every
single proof we've seen

1457
01:19:31,080 --> 01:19:32,496
and even the ones
we haven't seen.

1458
01:19:32,496 --> 01:19:35,000
You can always compute that
bijection between solutions

1459
01:19:35,000 --> 01:19:38,110
of A to solutions of B.
And so what that means is,

1460
01:19:38,110 --> 01:19:46,550
if I can solve B, consider
the ASP version of A

1461
01:19:46,550 --> 01:19:47,890
where I give you a solution.

1462
01:19:47,890 --> 01:19:50,610
If I can convert that solution
into a solution for my B

1463
01:19:50,610 --> 01:19:52,800
problem, and then with
the parsimonious reduction

1464
01:19:52,800 --> 01:19:58,830
I also get a B instance, then
I can solve the ASP problem

1465
01:19:58,830 --> 01:20:00,560
for B. That's a
decision problem.

1466
01:20:00,560 --> 01:20:02,000
Is there another solution?

1467
01:20:02,000 --> 01:20:04,480
If yes, then A will also
have another solution,

1468
01:20:04,480 --> 01:20:07,420
because it's parsimonious
the numbers will be the same.

1469
01:20:07,420 --> 01:20:11,050
We need a weaker version of
parsimony we just need the one

1470
01:20:11,050 --> 01:20:13,011
and more than one
are kept the same.

1471
01:20:13,011 --> 01:20:15,260
If this one was unique then
this one should be unique.

1472
01:20:15,260 --> 01:20:18,040
If this one wasn't unique then
this one should be not unique.

1473
01:20:18,040 --> 01:20:20,180
If I can solve B
then I can solve A.

1474
01:20:20,180 --> 01:20:24,070
Or if I can solve ASP B
then I can solve ASP A.

1475
01:20:24,070 --> 01:20:27,750
A lot of these proofs are
done-- so-called ASP reduction

1476
01:20:27,750 --> 01:20:30,660
is usually done by a
parsimonious reduction.

1477
01:20:30,660 --> 01:20:33,740
And so in particular
this was introduced,

1478
01:20:33,740 --> 01:20:36,410
this concept was introduced in
the likes of like Slitherlink,

1479
01:20:36,410 --> 01:20:40,160
I think, was one of the early
ASP completeness proofs.

1480
01:20:40,160 --> 01:20:42,310
And they were interested
in this because the idea

1481
01:20:42,310 --> 01:20:43,935
is if you're designing
a puzzle usually

1482
01:20:43,935 --> 01:20:46,390
you design a puzzle
with a solution in mind.

1483
01:20:46,390 --> 01:20:48,780
But then you need to make sure
there's no other solution.

1484
01:20:48,780 --> 01:20:52,380
So you have exactly
this set up and if you

1485
01:20:52,380 --> 01:20:55,020
want a formal definition of ASP
completeness it's in the notes.

1486
01:20:55,020 --> 01:21:00,430
But it's not that difficult.

1487
01:21:00,430 --> 01:21:03,602
The key point here is if you're
going to prove sharp P or ASP

1488
01:21:03,602 --> 01:21:06,060
completeness you might as well
prove the other one as well.

1489
01:21:06,060 --> 01:21:09,400
Get twice the results for
basically the same reduction.

1490
01:21:09,400 --> 01:21:11,460
All the versions of
3SAT and 1-in-3SAT,

1491
01:21:11,460 --> 01:21:14,790
and all those things, those
were all parsimonious.

1492
01:21:14,790 --> 01:21:17,730
And all of those are
ASP complete as well.

1493
01:21:17,730 --> 01:21:19,330
But once you get
into C-monious you're

1494
01:21:19,330 --> 01:21:23,470
no longer preserving one
versus more than one.

1495
01:21:23,470 --> 01:21:26,252
So while you preserved--
from a counting perspective

1496
01:21:26,252 --> 01:21:28,460
in multicall reductions
where you can divide that off

1497
01:21:28,460 --> 01:21:29,320
that's fine.

1498
01:21:29,320 --> 01:21:32,490
But from an ASP completeness
perspective you're not fine.

1499
01:21:32,490 --> 01:21:36,010
So in fact all the
stuff that we just

1500
01:21:36,010 --> 01:21:40,870
did with the permanent
matchings and sharp 2SAT,

1501
01:21:40,870 --> 01:21:41,880
their sharp P complete.

1502
01:21:41,880 --> 01:21:43,625
They may not be ASP complete.

1503
01:21:46,230 --> 01:21:48,080
But you'll notice
the puzzles that we

1504
01:21:48,080 --> 01:21:54,940
reduce from like, whatever,
Shakashaka is one of them.

1505
01:21:54,940 --> 01:21:59,210
That was from a version of
3SAT, and the other one we had

1506
01:21:59,210 --> 01:22:00,710
was from a version
of Hamiltonicity.

1507
01:22:03,500 --> 01:22:05,130
This guy.

1508
01:22:05,130 --> 01:22:07,090
These are all-- which
was also from 3SAT.

1509
01:22:07,090 --> 01:22:10,250
So these are all
ASP complete also.

1510
01:22:10,250 --> 01:22:12,850
But if you use the
weirder things like 2SAT

1511
01:22:12,850 --> 01:22:14,170
you don't get that.

1512
01:22:14,170 --> 01:22:17,070
So usually in one proof you can
get NP hardness, ASP hardness,

1513
01:22:17,070 --> 01:22:19,127
and sharp P hardness.

1514
01:22:19,127 --> 01:22:20,960
But if you're going to
go from these weirder

1515
01:22:20,960 --> 01:22:22,777
problems with C-monious
reductions then

1516
01:22:22,777 --> 01:22:24,860
you only get sharp P
hardness, because they're not

1517
01:22:24,860 --> 01:22:26,900
even NP-hard.

1518
01:22:26,900 --> 01:22:28,230
Cool, all right.

1519
01:22:28,230 --> 01:22:30,000
Thank you.