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ERIC DEMAINE: All right,
today we do NP completeness,
an entire field in one lecture.
Should be fun.
I actually taught an entire
class about this topic
last semester, but now we're
going to do it in 80 minutes.
And we're going to look at
lots of different problems,
from Super Mario Brothers
to jigsaw puzzles,
and show that
they're NP -complete.
This is a fun area.
As Srini mentioned last class,
it's all about reductions.
It's all about
converting one problem
into another, which is a fun
kind of puzzle in itself.
It's an algorithmic challenge.
And we're going to do it a lot.
But first I'm going to remind
you of some of the things you
learned from 006, and tell you
what we need to do in order
to prove all of these relations,
what exactly we need to show
for each of those arrows,
and why it's interesting.
So this is generally around
the P versus NP problem.
So remember, P is all
the problems we know how
to solve in polynomial time.
Well not just the ones we
know how to solve, but also
the ones that can
be solved, which
is pretty much-- which is the
topic of 6.006, and 6.046 up
till now.
But for now, in the
next few lectures,
we'll be talking about
problems that are probably not
polynomially solvable,
and what to do about them.
Polynomial, as you now, is
like n to the some constant.
Polynomial good exponential bad.
What is n?
I guess n is the size
of the problem, which
we'll have to be a little
bit careful about today.
And then NP is not
problem solvable
not in polynomial time,
but it's problem solvable
in nondeterministic
polynomial time.
And in this case
we need to focus
on a particular type of problem,
which is decision problems.
Decision just means that the
answer is either yes or no.
So it's a single bit answer.
We will see why we
need to restrict
to that kind of
problem in a moment.
So this is problems you can
solve in polynomial time.
Same notion of polynomials,
same notion of n,
but in a totally unrealistic
model of computation.
Which is a
nondeterministic model.
In a nondeterministic
model, what
you can do is say
instead of computing
something from something you
know, you could make a guess.
So you can guess one out of
polynomially many options
in constant time.
So normally a constant time
operation, in regular models,
like you add two numbers, or you
do an if, that sort of thing.
Here we can make a guess.
I give the computer polynomially
many options I'm interested in.
Computer's going to
give me one of them.
It's going to give
me a good guess.
Guess is guaranteed to be good.
And good means here
that I want to get
to a yes answer if I can.
So the formal statement
is, if any guess
would lead to a yes answer,
then we get such a guess.
OK, this is weird.
And it's asymmetric.
It's biased towards yes.
And this is why we can only
think about decision problems,
yes or no.
You could bias towards no.
You get something
else called coNP.
But we'll focus here just on NP.
So the idea is I'd really
like to find a guess that
leads to a yes answer.
And the machine magically
gives me one if there is one.
Which means if I end
up saying no, that
means there was absolutely no
path that would lead to a yes.
So when you get a no, you
get a lot of information.
When you get a yes, you
get some information.
But hey, you were lucky.
Hard to complain.
So in 006, I often call this
the lucky model of computation.
That's the informal version.
But nondeterminism is
what's really going on here.
So maybe it's useful
to get an example.
So here's a problem we'll--
this is sort of the granddaddy
of all NP-complete problems.
We'll get to
completeness in a moment.
3SAT-- SAT stands
for satisfiability.
So in 3SAT, the
input to the problem
looks something like this.
I'm just going to
give an example.
And in case you've forgotten
your weird logic notation,
this is an and.
These are ORs.
And I'm using this
for negation, not.
So in other words, I'm given
a formula which is and of ORs.
And each or clause only
has three things in it.
These things are
called literals.
And a literal is either
a variable x sub i,
or it's the negation of
a variable, not x sub i.
So this is a typical example.
You could have no negations.
You could here have one
negation, two negations,
any number of
negations per clause.
These groups of three--
these or of three
things, three literals,
are called clauses.
And they're all ANDed together.
And my goal is, this should
be a decision question,
so I have a yes or no question.
And that question is, can
you set the variables--
So they're x1 to true or false?
So each variable I get to choose
a true or false designation
such that the formula
comes out true.
I use T and F for
true and false.
So I want to set
these variables such
that every clause comes
out true, because they're
ANDed together.
So I have to satisfy this
clause in one of three ways.
Maybe I satisfy
it all three ways.
Doesn't matter, as long
as at least one of these
should be true, and at least
one of these should be true,
and at least one of each
clause should be true.
So that's the 3SAT problem.
This is a hard problem.
We don't know a
polynomial time algorithm.
There probably isn't one.
But there is a polynomial time
nondeterministic algorithm.
So this problem is in NP
because if I have lucky guesses,
it's kind of designed to
solve this kind of problem.
What I'm going to do is guess
whether x1 is true or false.
So I have two choices.
And I'm going to ask my machine
to make the right choice,
whether it should
be true or false.
Then I'll guess x2.
Each of these guess operations
takes constant time.
So I do it for every variable.
And then I'm going
to check whether I
happen to satisfy the formula.
And if it comes out true,
then I'll return yes.
And if it comes out
false, I'll return no.
And because NP is biased
towards yes answers,
it always finds a yes
answer if you can.
If there's some way to
satisfy the formula,
then I will get it.
If there's some way to make
the formula come out true,
then this algorithm
will return yes.
If there's no way
to satisfy it, then
this nondeterministic
algorithm will return no.
That's just the
definition of how
nondeterministic machines work.
It's a little weird.
But you can see from
this kind of prototype
of a nondeterministic algorithm,
you can actually always
arrange for your guessing
to be at the beginning.
And then you do some regular
polynomial time checking
or deterministic checking.
So when you rewrite
your algorithm
like this with guesses up
front and then checking,
you can also think of it as
a verification algorithm.
So you can say, your friend
claims that this 3SAT formula
is satisfiable,
meaning there's a way
to set the variable so
that it comes out true.
So this is called a
satisfying assignment.
Satisfying just means make true.
And you're like, no,
I don't believe you.
And your friend says no,
no, no, really, it's true.
And here's how I can prove it.
You set x1 to false.
You set x2 to true.
You set x3-- basically
they give you the guesses.
And then you don't
have to be convinced
that those are
the right guesses,
you can check that
it's the right guess.
You can compute this
formula in linear time,
see what the outcome is.
If someone tells you
what the xi's are,
you can very quickly
see whether that
was a satisfying assignment.
So you could call
this a solution,
and then there's a
polynomial time verification
algorithm that checks
that solutions are valid.
But, you can only do
that for yes answers.
Your friend says no,
this is not satisfiable,
they have no way of
proving it to you.
I mean, other than checking
all the assignments separately,
which would take
exponential time,
there's no easy way to confirm
that the answer to this problem
is no.
But there is an
easy way to check
that the answer is yes, namely
I give you the satisfying
assignment.
So this definition of NP
is what I'll stick to.
It's this sort of-- I
like guessing because it's
like dynamic programming.
With dynamic programming
we also guess,
and guessing actually originally
comes from this world,
nondeterminism.
In dynamic programming, we
don't allow this kind of model.
And so we have to check
the guesses separately.
And so we spend lots of time.
Here, magically, you
always get the right
guess in only constant time.
So this is a much
more powerful model.
Of course there's no computers
that work like this, sadly,
or I guess more interestingly.
So this is more about confirming
that your problem is not
totally impossible.
At least you can check the
answers in polynomial time.
So that's one thing.
So this is an equivalent
definition of NP
because you can take a
nondeterministic algorithm
and put the guessing up top.
You can call the
results of those guesses
a certificate that
an answer is yes.
And then you have a regular old
deterministic polynomial time
algorithm that, given
that certificate,
will verify that it actually
proves that the answer is yes.
It's just that certificate
has to be polynomial size.
You can't guess something
of exponential size.
You can only guess something of
polynomial size in this model.
So seems a little weird.
But we'll see why this is
useful in a little bit.
So let me go to NP completeness.
So if I have a problem X,
it's NP-complete if X is in NP
and X is NP-hard.
But I haven't told
you what NP-hard is.
Maybe you remember from
006, but let me remind you.
So, I need to define reduce.
So maybe I'll do
that as well, then
we can talk about all these.
OK, a lot of definitions.
But the idea of NP
hardness is very simple.
If problem X is
NP-hard, it means
that it's at least
as hard as-- sorry,
that is a Y-- it's at least
as hard as all problems in NP.
Intuitively, X means
it's at least as
hard as everything in NP.
Whereas being in NP is
a positive statement.
That says it's not
too hard, at least
there's a polynomial time
verification algorithm.
So being in NP is good news.
It says you're no
harder than NP.
NP-hard says you're at least
as hard as everything in NP.
And so NP-complete
is a nice answer
because this says you're exactly
as hard as everything in NP--
no harder, no easier.
If you draw, in
this vague sense,
computational difficulty
on one axis-- which is not
really accurate, but I
like to do it anyway--
and you have P is all of
these easy problems down here.
And NP is some
larger set like this.
NP-hard is from here over.
And this point right
here is NP-complete.
Being in NP means you're left
of this line, or on the line.
And being NP-hard means
you're right of this line,
or on the line.
NP-complete means
you're right there.
So that's a very definitive
sense of hardness.
Now there is this
slight catch, which
is we don't know
whether P equals NP.
So maybe this is the same
as this, but probably not.
Unless you believe
in luck, basically,
unless you imagine
that a computer could
engineer luck and always
guess the right things
without spending a lot of
time, then P does not equal NP.
And in that world, what
we get is that if you have
an NP-complete problem, or
actually any NP-hard problem,
you know it cannot be NP.
So if you have
that X is NP-hard,
then you know that X is
not in P unless all of NP
is in P. So unless P equals NP.
And most reasonable people
do not believe this.
And so instead they
have to believe this,
that your problem is not
polynomially solvable.
So why is this true?
Because if your
problem is NP-hard,
it is at least as hard
as every problem in NP.
And if you believe that
there is some problem in NP--
we don't necessarily
know which one--
but if there is any problem out
there in NP that is not in P,
then X has to be at
least as hard as it.
So it also requires
nonpolynomial time, something
larger than polynomial time.
What does at least
as hard mean though?
We're going to define it
in terms of reductions.
Reduction from one
problem to another
is just a polynomial
time algorithm,
regular deterministic
polynomial time,
that converts an
input to the problem A
into an equivalent
input to problem
B. Equivalent means that it
has the same yes or no answer.
And we'll just be thinking
about decision problems today.
So why would I care
about a reduction?
Because what it tells me is that
if I know how to solve problem
B, then I also know how
to solve problem A. If I
have a, say, a polynomial
time algorithm for solving B
and I want one for A,
I just take my A input.
I convert it into the
equivalent B input.
Then I run my algorithm
for B, and then it
gives me the answer
to the A problem
because the answers
are the same.
So if you have a reduction
like this and if say,
B, has a polynomial
time algorithm,
then so does A, because you
can just convert A into B,
and then solve B.
Also this works
for nondeterministic algorithms.
Not too important.
So what this tells us is
that in a certain sense--
get this right--
well this is saying,
if I can solve B,
that I can solve A.
So this is saying
that B is at least as
hard as A. I think I got
that right, a little tricky.
So if we want to prove
the problem is NP hard,
what we do is show that
every problem in NP
can be reduced to the problem
of X. So now we can go back
and say well, if we believe
that there is some problem Y,
that is in NP minus P, if
there's something out here that
is not in P, then we
can take that problem Y,
and by this definition,
we can reduce it
to X, because everything
in NP reduces to X.
And so then I can
solve my problem
Y, which is in NP minus P,
by converting it
to X and solving
X. So that means X better
not have a polynomial time
algorithm, because if
it did, Y would also
have a polynomial
time algorithm.
And then in general,
P would equal
NP, because every problem
in NP can be converted to X.
So if X has a polynomial
time algorithm,
then every problem Y does.
Question?
AUDIENCE: For the
second if statement,
why can't you say that if
A is in NP, B is in NP?
ERIC DEMAINE: So you're asked
us about the reverse question.
If is A in NP, can we
conclude that B is in NP?
And the answer is no.
Because this reduction only
lets us convert from A to B.
It doesn't let us do anything
for converting from B to A.
So if we know how to
solve A and we also
know how to convert A into B,
it doesn't tell us anything.
It could be B is a much
harder problem than A,
in that situation.
That's, I think, as good
as I can do for that.
Other questions?
All right.
It is really tricky to get
these directions right.
So let me give you a handy guide
on how to not make a mistake.
So maybe over here.
What we care about, from
an algorithmic perspective,
is proving the problems
are NP-complete.
Because if we prove
NP-completeness-- I mean,
really we care
about NP-hardness,
but we might as well
do NP-completeness.
Most of the problems that
we'll see that are NP-hard
are also NP-complete.
So when we prove this, we
prove that there is basically
no polynomial time
algorithm for that problem.
So that's good to
know, because then we
can just give up searching for
a polynomial time algorithm.
So all the problems
we've seen so far
have polynomial time algorithms,
except a couple in your problem
sets, which were
actually NP-complete.
And the best you could
have done was exponential,
unless P equals NP.
So here's how you can prove
this kind of lower bound
to say look, I don't need
to look for algorithms
any more because my
problem is just too hard.
It's as hard as
everything in NP.
So this is just a summary
of those definitions.
The first thing you do
is prove that X is in NP.
The second thing you do is
prove that X is NP-hard.
And to do that, you reduce
from some known NP-complete
problem-- or I guess
NP-hard, but we'll
use NP-complete--
to your problem
X. Maybe I'll give
this a name Y.
OK, so to prove
that X is in NP, you
do something like
what we did over here,
which is to give a
nondeterministic algorithm.
Or you can think
of it as defining
what the certificate is and
then giving a polynomial time
verification algorithm.
So sort of two approaches.
You can give a nondeterministic
polynomial time algorithm,
or you give a certificate
and a verifier.
There's no right or
wrong certificate.
I mean, a certificate, you
can define however you want,
as long as the verifier
can actually check it
and when it says yes, then the
answer to the problem was yes.
So it's really the same thing.
Just want to say
there's some certificate
that a verifier
could actually check.
So that's proving that
your problem is in NP.
It's sort of an
algorithmic thing.
The second part is
all about reductions.
Now the definition
says that I should
reduce every problem
in NP to my problem X.
That's tedious,
because there are
a lot of problems in the world.
So I don't want to do it
for every problem in NP.
I'd like to just do it for one.
Now if I reduce
sorting to my problem,
that's not very interesting.
It says my problem is at
least as hard as sorting.
But I already know
how to solve sorting.
But if I start from an
NP-complete problem,
then I know, by the definition,
that every problem in NP
can be reduced to that problem.
And if I show how to reduce
the NP-complete problem to me,
then I know that
I'm NP-complete too.
Because if I have
any problem Z in NP,
by the definition of NP-complete
of Y I can reduce that to Y.
And then if I can build
a reduction from Y to X,
then I get this reduction.
And so that means I can
convert any problem in NP
to my problem X, which
means X is NP-hard.
That's the definition.
So all this is to
say the first time
you prove a problem is
NP-complete in the world-- this
happened in the '70s by Cook.
Basically he proved that
3SAT is NP-complete.
That was annoying,
because he had
to start from any
problem in NP, and he
had to show that you could
reduce any problem in NP
to 3SAT.
But now that that hard work is
done, our life is much easier.
And in this class all
you need to think about
is picking your favorite
NP-complete problem.
3SAT's a good choice
for almost anything,
but we'll see a bunch of other
problems today from here.
And then reduce from that
known problem to your problem
that you're trying
to prove is NP-hard.
If you can do that, you know
your problem is NP-hard.
So we only need one reduction
for each hardness result, which
is nice.
And this picture is a
collection of reductions.
We're going to start from 3SAT.
I'm not going to prove
that it's NP-complete,
although I'll give you a
hint as to why that's true.
We're going to reduce it
to Super Mario Brothers.
We're going to reduce it to
three dimensional matching.
We're going to reduce
three dimensional matching
to subsets sum, to partition,
to rectangle packing,
to jig saw puzzles.
And we're going to do all
those reductions, hopefully.
And that's proving NP-hardness
of all those problems.
They're also all in NP.
So 30 second intuition
why 3SAT is NP-hard.
Well, if you have
any problem in NP,
that means there is one of these
nondeterministic polynomial
time algorithms, or there
is some verifier given
a polynomial size certificate.
So that verifier is
just some algorithm.
And software and hardware
are basically the same thing,
right?
So you can convert
that algorithm
into a circuit that
implements the algorithm.
And if I have a circuit with
like ANDs and ORs and NOTs,
I can convert that
into a Boolean formula
with ANDs, ORs, and NOTs.
Circuits and formulas
are about the same.
And if I have a
formula-- fun fact,
although this is a
little less obvious--
you can convert it into this
form, an AND of triple ORs.
And once you've done
that, that formula
is equivalent to the
original algorithm.
And the inputs to that
verification algorithm,
the certificate, are represented
by these variables, the xi's.
And so deciding whether
there's some way
to set the xi's to
make the formula true
is the same thing as saying is
there some certificate where
the verifier says yes, which
is the same thing as saying
that the problem has answer yes.
So given an NP algorithm, one
of these nondeterministic funny
algorithms, we can convert it
into a formula satisfaction
problem.
And that's how you prove
3SAT is NP-complete.
But to do that can
take many lectures,
so I'm not going
to do the details.
The main annoying part is
being formal about what exactly
an algorithm is, which we
don't do in this class.
If you're interested,
take 6.045,
which is some
people are actually
in the overlap this semester.
Cool.
Let's do some reductions.
This is where things get fun.
So we're going to start
with reducing 3SAT
to Super Mario Brothers.
So how many people have
played Super Mario Brothers?
Easy one.
I hope if you haven't
played, you've seen it,
because we're going to rely very
much on Super Mario Brothers
physics, which I hope
is fairly intuitive.
But if you haven't played,
you should, obviously.
And we're going to reduce
3SAT to Super Mario Brothers.
Now this is a theorem by a
bunch of people, one MIT grad
student, myself, and a couple
other collaborators not at MIT.
And of course this result holds
for all versions of Super Mario
Brothers so far
released, I think.
The proofs are a
little bit different
for each one, especially Mario
2, which is its own universe.
What I'm going to talk about the
original Super Mario Brothers,
NES classic which
I grew up with.
Now the real Super Mario
Brothers is on a 320
by 240 screen.
It's a little bit small.
Once you go right, you can't go
back left, except in the maze
levels anyway.
So I need to generalize
a little bit.
Because if you assume that
the screen size of Super Mario
Brothers is
constant, in fact you
can dynamic program
your way through
and find the optimal
solution in polynomial time.
So I need to generalize a little
bit to arbitrary board size,
arbitrary screen size.
So in fact, my entire level will
be in one screen, no scrolling.
Never mind this is a
side scrolling adventure.
And so that's my
generalized problem.
And I claim this is NP-hard.
If I give you a
level and I ask you,
can you get to the
end of this level?
That problem is NP-hard.
Also no time limit.
The time limit would be OK,
but you have to generalize it.
Instead of 300
seconds or whatever,
it has to be an arbitrary value.
So how are we going to do this?
We're going to reduce from
3SAT to Super Mario Brothers.
So that means I'm given--
I don't get to choose.
I'm given one of these formulas.
And I have to convert it into an
equivalent Super Mario Brother
instance.
So I have to convert it into
a level, a hypothetical level
of Super Mario Brothers.
Given a formula, I have
to build a level that
implements that formula.
So here's what it's
going to look like.
I'm going to start
out somewhere.
Here's my drawing of Mario.
Mario-- or you could play Luigi.
It doesn't matter.
First thing it's going to do
is enter a little black box
called a variable.
This is supposed to
represent, let's call it x1.
And so it's some black box.
I'm going to tell you
what it is in a moment.
And it has two outputs.
There's the true output
and the false output.
And the idea is that Mario has
to choose whether to set x1
to true or false.
Let me show you that gadget.
So here's the-- whoops,
upside down-- here
is the variable gadget.
So here's Mario.
Could enter from
this way or that way.
We'll need a couple of
entrances in a moment.
And then falls down.
Once Mario is down here, if
you check the jump height,
you cannot get back up to here.
So this is like a one way.
Once you're down here,
you have a choice.
Should I fall to the left
or fall to the right?
And if you make these falls
large enough, once you fall,
you can't unfall.
So once you make a
choice of whether I
leave on the true exit
or the false exit,
that's a permanent choice.
So you can't undo it, unless
you can come back to here.
But we'll set up so
that never happens.
I mean, if you're trying
to solve the level,
you don't know which way to go.
You have to guess.
Can I go fall to the left
or fall to the right,
or do something.
So the existence of a
play through, this level,
is the same as saying there is
a choice for the x1 variable.
Now we have to do this
for lots of variables.
So there's x2 variable,
x3 variable, and so on.
Each one has a true
exit and a false exit.
So the actual level will
have n instances of this
if we have n variables.
Now, what do I do
once Mario decides
that this is a true thing?
What I'm going to
do is have-- this
is called a gadget by the way.
In general, most
NP-hardness proofs
use these things called
gadgets, which is just saying,
we take various
features of the input,
and we convert them into
corresponding features
on the output.
So here I'm taking each
variable, x1, x2, x3,
and so on, and building
this little gadget
for each of those variables.
Now the other main thing you
have in 3SAT are the clauses.
We have triples of variables
or their negations.
They have to come
together and be satisfied.
One of them has to be true.
So down here I'm going to have
some clause gadgets, which
I will show you in a moment.
OK, and I think
I'll switch colors.
This is about to get messy.
So the idea is that some of
the clauses have x1 in them.
The true version
of x1, not x1 bar.
So for those clauses,
I want to connect.
I'm going to dip into
the clause briefly.
So from this wire going to
dip into the clause here.
And then I'm going to go to
the next clause that has x1.
Maybe it's this one, and
the next one, and so on.
All the clauses that have
x1 in it, I dip into.
The other ones I don't.
And then once I'm
done, I'm going
to come back and feed into x2.
Next, I look at this
false wire for x1.
So all the clauses that
have x1 bar in them,
I'm going to connect.
So I don't know
which ones they are.
Maybe this one, or
this one, something.
And then I come here.
And so the idea is that
Mario makes a choice
whether x1 is true or false.
If x1 is true, Mario is going
to visit all of the clauses that
have x1 true in them.
And then it's going to
go to the x2 choice.
Then it's going to choose
whether x2 is true or false,
and repeat.
Or Mario decides
x1 should be false.
That will satisfy
all the clauses
that have x1 bar in them.
And then again, we
feed back into x2.
So this is why we have two
inputs into the x2 gadget.
One of them is when the
previous variable was true.
The other is when the
previous variable was false.
The choice of x2 doesn't
depend on the choice of x1.
So they feed into
the same thing.
And you have to
make your choice.
So far, so good.
Now the question is, what's
happening in these clauses.
And then there's
one other aspect,
which is after you've set all of
the variables, at the very end,
after this last variable
xn, at the very end,
what we're going to do is come
and go through all the clauses.
And then this is the flag.
This is where you win the level.
Sorry, I drew it backwards.
But the goal is for Martin to
start here and get to here.
In order to do
that, you have to be
able to traverse
through these clauses.
So what do the
clauses look like?
This is a little
bit more elaborate.
So here we are.
This is a clause gadget.
So there are three ways
to dip into the clause.
It's actually upside down
relative to that picture,
but that's not a problem.
So if Mario comes here, then
he can hit the question mark
from below.
And inside this question mark
is an invincibility star.
And the invincibility star
will come up here and just
bounce around forever.
We checked.
The star will just stay there
for as long as you let it sit.
Unfortunately, all of
these are solid blocks,
so Mario can't actually get
up to here to get the star.
But as long as Mario
can visit this question
mark or this question mark
or this question mark,
then there will be at
least one star up here.
So the idea is
that each of these
represents one of the
literals that's in the clause.
And if we choose-- so let's look
at this first clause, x1 or x3
or x6 bar.
So if we choose x1 to be true,
then we'll follow the path
and we'll be able
to hit the star.
Or if we choose x3 to be
true, then we'll come in here
and hit this star.
Or if we choose x6 to
be false, then that path
will lead to here and we'll be
able to hit this question mark
and get the star up here.
So as long as we
satisfy the clause,
there will be at least one star.
Won't help if you
have multiple stars.
Then the final traversal part--
so that was this first clause.
And now we're
traversing through.
Actually in this picture,
it's left to right.
Just turn your head.
And so now Mario
is going to have
to traverse this gadget from
left to right on this top part.
And if Mario comes in here and
you can barely jump over that.
If there's a star, you
can collect the star
and then run through all of
these flaming bars of death.
If there's no star, you can't.
You'll die if you
try to traverse.
So in order to be able to
traverse all these clauses,
they must all be true.
And them all being true is the
same is their AND being true.
So you will be able to survive
through all these clauses
if and only if this formula
has a satisfying assignment.
The satisfying
assignment would be
given to you by the level play.
The choices that Mario
makes in this gadget
will tell you
whether each variable
should be true or false.
So to elaborate just
a little bit more
in general, when you have
a reduction like this,
to prove that it actually works,
you need to check two things.
You need to check
that if there is a way
to satisfy this
formula, then there
is a way to play this level.
And then conversely you need
to show that if there's a way
to play this level,
then the formula
has a satisfying assignment.
So for that latter
part, in order
to convert a level play into
a satisfying assignment,
you just check which way Mario
falls in each of these gadgets,
left or right.
That tells you the
variable assignment.
And because of the
way the clauses work,
you'll only be able to
finish the level if there
was at least one star here.
And stars run out
after some time.
So you can barely make it
through all the flaming bars
of death.
Then you get to the next clause.
You need another
star for each one.
Conversely, if there is
a satisfying assignment,
you can actually play
through the level,
you just make these
choices according to what
the satisfying assignment is.
So either way it's equivalent.
We always get a yes or
no answer here whenever
we get a corresponding yes or
no answer to the 3SAT process.
You also need to check that this
reduction is polynomial size.
It can be computed
in polynomial time.
So there's an issue.
Given this thing, you have
to lay this out in a grid
and draw all these wires.
And there's one
problem here, which is,
these wires cross each other.
And that's a little
awkward, because these wires
are basically just long tunnels
for Mario to walk through.
But what does it mean
to have a crossing wire?
Really, if Mario's
coming this way,
I don't want them to
be able to go up here.
He has to go straight.
Otherwise this
reduction won't work.
So I need what's called
a crossover gadget.
And everywhere here I have a
crossing, I have a crossover.
And this gadget has
to guarantee that I
can go through one
way or the other way,
but there's no leakage from
one path to the other path.
Actually, if I first
traverse through here,
and then I traverse through
here, it's OK if I leak back.
Because once I visit a
wire, it's kind of done.
But I can't have leakage if
only one of them is traversed.
So this is the last gadget, the
most complicated of them all.
So this took a
while to construct,
as you might imagine.
So this is what we call a
unidirectional crossover.
You can either go from left to
right or from bottom to top,
but you cannot go from bottom to
right or bottom to left or left
to bottom, that kind of thing.
So I'm told that
Mario is only going
to enter from here to here,
because all of these wires,
I can make one way wires.
I only have to think about
going in a particular direction.
I can have falls to
force Mario to only go
one way along these wires.
And so let me show you
the valid traversals.
Maybe the simplest
one is from here.
So let's say Mario
comes in here, falls.
So I can't backtrack,
can jump up here.
And then if Mario's big,
he can break this block,
break this block.
But if he's big-- there should
be a couple more zig zags here.
Let's try to run.
You can crouch
slide through here.
But then you'll sort
of lose your momentum,
and you won't be able
to go through all
these traversals as big Mario.
So you can break these blocks
and then get up to the top
and leave.
Or, if big Mario comes
from over this way,
you can first take a
damage, become small Mario.
Then you can fit through
these wiggly blocks.
But you cannot break blocks
anymore as small Mario.
So once you've committed
to going small,
you have to stay small,
until you get to here.
And then there's a
mushroom in this block.
So you can get big again, and
then you can break this block
and leave.
But once you're big,
you can't backtrack
because big Mario can't fit
through these tiny tubes.
See it clear, right?
So slight detail, which
is at the beginning,
we need to make Mario big--
so there's a little mushroom.
I think they have three
spots-- at the beginning.
And also at the
end, there has to be
something like this that
checks that you actually
have a mushroom.
So the only time you're
allowed to take damage
is briefly in this
gadget you take damage.
If you tried to backtrack,
you would get stuck.
There's a long fall here.
And then you have
to get the mushroom
so you can escape again.
So at the end there's
like a mushroom check.
Make sure you have it.
So most of the
time Mario is big.
And just in these
little crossovers
you have to make
these decisions.
This would make a
giant level, but it
is polynomial size, probably
quadratic or something.
Therefore Super Mario
Brothers is NP-hard.
So if you want more
fun examples like this,
you should check out 6.890, the
class I taught last semester,
which has online video lectures,
soon to be on OpenCourseWare.
So you can play with that.
Any questions about Mario?
All right, I hope you all play.
So the next topic is
a problem you probably
haven't heard about, three
dimensional matching.
This is a kind of a
graph theory problem.
We're going to call
it 3DM for short.
And you've seen matching
problems based on flow.
Matching problems are usually
about pairs of things.
You're pairing them
up, which you might
call two dimensional matching.
That can be solved
in polynomial time.
But if you change two to three
and you're tripling things up,
then suddenly the problem
becomes NP-complete.
So it's a useful starting
point, similar to 3SAT.
So you're given a
set X of elements,
a set Y of elements,
a set Z of elements.
None of them are shared.
But more importantly, you
are given a bunch of triples.
These are the allowable triples.
So we'll call the set
of allowable triples T.
And so we're looking
at the cross product.
This is the set of all triples
X, Y, and Z, or X is in X,
Y is in Y, and Z is in Z. But
not all triples are allowed.
Only some subset of
triples is allowed.
And your goal is to choose
among those subsets-- sorry,
among those triples a
subset of the triples.
So we're trying
to choose a subset
S of T such that every element--
so the things in X, Y, and Z
are called elements.
So I'm just taking
somebody in the union XYZ.
It should be in exactly
one triple s in big S.
This is a little weird, but
you can think of this problem
as you have an alien
race with three genders--
male, female, neuter I guess.
Those are the X, Y, and Z's.
There's an equal number of each.
And every triple reports
to you whether that
is a compatible matching.
Who knows what they're
doing, all three of them?
So you're told up front--
you take a survey.
There's only n cubed
different triples.
For each of them they
say, yeah, I'd do that.
So you were given that subset.
And now your goal is to
permanently triple up
these guys.
And everybody wants to
be in exactly one triple.
So it's a monogamous
race, imagine.
So everybody wants to be put
in one triple, but only one
triple.
And the question is,
is this possible?
This is three
dimensional matching.
Certainly not always going to be
possible, but sometimes it is.
If it is, you want
to answer yes.
If it's not possible,
you want to answer no.
This problem is NP-complete.
Why is it in NP?
Because I can basically guess
which elements of T are in S.
There's only at most
n cubed of them.
So for each one, it
is guess yes or no,
is that element of T in S?
And then I check whether this
coverage constraint holds.
So it's very easy to
prove this is in NP.
The challenge is to
prove that it's NP-hard.
And we're going to do that,
again, by reducing from 3SAT.
So we're going to make
a reduction from 3SAT
to three dimensional matching.
Direction is important.
Always reduce from the
thing you know is hard
and reduce to the thing
you don't know is hard.
So again, we're given a formula.
And we want to
convert that formula
into an equivalent three
dimensional matching input.
So the formula has
variables and clauses.
For each variable,
we're going to build
a gadget that looks like this.
And for each clause we're
going to build a gadget.
So here's what they look like.
If we have a variable x1,
we're going to convert that
into this picture.
Stay monochromatic for now.
Looks pretty crazy at the
moment, but it's not so crazy.
This is not supposed
to be obvious.
You have to think for a while.
It's a puzzle to figure
out this kind of thing.
But I call this thing a variable
gadget because locally--
so there's basically a
wheel in the center here.
And then there's
these extra dots
for every pair of dots,
consecutive pairs of dots
in a wheel.
And what I've drawn is the set
of triples that are allowed.
There's tons of other
triples which are forbidden.
The triples that are
in T are the ones
that I draw as little triangles.
And two color them because
there are exactly two ways
to solve this gadget locally.
Now these dots are going to
be connected to other gadgets.
But these dots only exist
in this gadget, which means
they've got to be covered.
They've got to be
covered exactly once.
So either you choose
the blue triangles,
or you choose the red triangles.
Each of them will exactly
cover each of these guys once.
You cannot mix and
match red and blue,
because you either get overlap
if you choose two guys that
share a point, or
you'd miss one.
If I choose like this
blue and this red,
then I can't cover this
point because both of these
would overlap those two.
And over here you have to
choose [INAUDIBLE] triples.
They can't overlap at all.
And everybody has
to get covered.
So just given those
constraints, locally you
can see you have to
choose red or blue.
Guess what?
One of them is true,
the other one is false.
Let's say that red is
true and blue is false.
In general, when you're trying
to build a variable gadget,
you build something
that has exactly
two solutions, one representing
true, one representing false.
Now how big do I
make this wheel?
Big enough.
You could make it as big
as the number of clauses.
I'm going to make
it into two and x1.
So wheel-- and this number
is the number of occurrences
of x1 in the formula.
So this is the number of clauses
that contain either xi or xi
bar.
That's in xi.
I'm going to double that.
Because what I get over
here is basically xi
being true for those guys.
Actually, yeah,
that's actually right.
It looks backwards.
And false for these guys.
One way or the other,
we'll figure it out.
So in order for xi to appear
in, say, five different clauses,
I want five of the true things
and five of the false things.
And so I need to double in
order to get-- potentially
I have twice as many
as I actually need,
but this way I'm guaranteed
to have false or true,
whichever I need.
In reality I have
some true occurrences.
I have some false occurrences,
some x1's, some x1 bars.
This will guarantee that I have
enough of these free points
to connect into
my clause gadgets.
How do I do a clause gadget?
It's actually really easy.
So these would be pretty
boring by themselves.
So a clause always
looks like this.
Maybe there's some negations.
Yeah, let's do
something like that.
I'm going to convert it
into a very simple picture.
It's going to be xi dot,
and xj bar dot, and xk dot.
And then-- well maybe I'll
stick to these colors.
Again, these two points only
appear in this clause gadget.
These dots are
actually these dots.
So there's one of
these pictures for x1.
There's another one for x2, x3.
And so xi has one
of these wheels.
I want this dot to be one
of these dots of the wheel.
And then I want
this dot to be one
of the dots in the xj wheel
with the false setting, one
of the red dots.
I want this one to be xk
true setting in the xk wheel.
So these things are
all connected together
in a complicated pattern.
But the point is that
within this gadget,
I only have three
allowed triples.
And these points only
appear in this gadget,
which means they have to
be covered in this gadget.
They can be covered
by this triple
or this triple or this triple.
But once you choose one,
you can't choose the others.
What this means is if
I set x1 to be true,
it leaves behind these
points marked true.
If I choose the red
things, then it's
the blue points that
are left behind.
Leaving points
behind in this case
is going to be good,
because this clause,
in order to satisfy
this clause, in order
to choose one of these three
triples, at least one of these
must be left behind
by the wheel.
If all of these are
covered by their wheels,
then there's no way.
I can't choose
any of these guys.
But if at least one of these
is left behind by the wheel,
then I can choose the
corresponding triple
and cover these points.
So I'll be able to cover
these points if and only
if at least one
of these is true.
And that's a clause.
That's what a clause is
supposed to do in 3SAT.
If at least one
of these is true,
then the clause is satisfied.
I need all the clauses
to be satisfied
because I need to cover
of these points for all
the instances of these clauses.
And that's how it works.
Now, slight catch.
If you do this,
not all the points
will be covered, even so.
Maybe all of these are true.
And so they're all left behind.
And I can only cover one
of them with the clause.
It's a little messy.
You need another gadget, which
is called garbage collection.
I don't want to spend
too much time on it.
But you have two dots.
And then you have every
single xi-- these dots,
all true and false dots.
And you're going to
have this triple,
and this triple, and this
triple, and this triple,
and so on.
It looks an awful
lot like a clause.
But this is like a clause
that's connected to everybody
in the entire universe.
And you repeat this
the appropriate number
of times, which is
something like sum of nx
minus the number of clauses.
OK, why?
Because if you
look at a wheel, it
has size 2 times nx
for a variable x.
And half of the points
will be left uncovered.
So that means nx of
them will be uncovered.
Then the clause, if everything
works out correctly,
the clause will cover
exactly one of those points.
So for each clause we
cover one of the points.
That means this
difference is exactly how
many points are left uncovered.
And so we make this gadget
exactly that many times.
And it's free to cover anybody.
So whatever is left over,
this garbage collector
will clean up.
And if we use exactly
the right number of them,
this garbage collector won't
run out of things to collect.
So this makes the proof messy.
But I want to move on to
somewhat simpler proofs
and for other problems.
Yeah?
AUDIENCE: Real quick,
what about the t or f
points that we didn't cover
because we didn't actually
need that many?
ERIC DEMAINE: Right.
So this also includes the points
that weren't even connected
to clauses.
I think this is the right
number no matter what,
because this is counting the
total number of uncovered guys,
whether they're connected
to clauses or not.
Each clause will, in
a satisfied situation,
it will cover exactly
one of those points.
The ones that are
connected to the clauses
won't be covered at
all, but that will still
be in this difference.
So yeah, it's good
to check that.
The first time I wrote this
down I forgot about those points
and got it wrong.
But I think this is
right, hopefully.
I did not come up
with this proof.
Garey and Johnson
I think-- or no.
This is-- I forgot.
Yeah, this is a Garey
and Johnson proof.
There's a cool book from the
late '70s by Garey and Johnson,
does a lot of NP-completeness,
if you're curious.
All right, so
hopefully you believe
three dimensional
matching is hard.
Now I'm going to use it to prove
that some very different types
of problems are hard.
This is a kind of
graph theory problem.
You'll see more graph theory
problems in recitation.
This one, I can
erase 3SAT and Mario.
So in the world, most
NP-hardness proofs
are reductions from 3SAT,
or some variation of 3SAT.
In some sense, you can think
of three dimensional matching
as kind of like a
version of 3SAT,
but it's a little
bit more stringent.
And that stringency helps
us to do other reductions.
So here's another
problem where we'll
reduce from three
dimensional matching.
It's called subset sum.
So you're given n
integers, a1 up to an.
And you're given a target
sum, also an integer.
Call it t.
What you'd like to know is, is
there a subset of the integers
that adds up to that target.
Can you choose a
sum of the integers
so that-- I'll write
it the sum of S.
But what this means is
the sum over the ai's that
are in S of the value ai.
I want that to equal t.
So this is the definition.
This is the constraint.
So I give you a
bunch of numbers.
Do any subset of
them add up to t?
That's all this is asking.
This problem is NP-hard.
It's NP-complete, in fact, when
you can guess which integers
should go in the subset,
and then add them up
to see if you got it right.
It is NP-hard, but
it's something special
we call weakly NP-hard.
And why don't I come back to the
definition of that in a moment?
Let me first show you the proof.
It's actually really easy
now that we have this three
dimensional matching problem.
It's pretty cool.
So these numbers are
going to be huge.
What we're going to say
is, let's view-- so again,
we're given a three
dimensional matching instance.
Get the directions, right?
We're given a set of triples.
We want to solve this problem
by reducing it to a subset sum.
So we get to construct integers
that represent triples.
That's what we're going to do.
So here we go.
We get to choose a number.
So I'm going to think of
them in a particular base, b,
which is going to be 1
plus the max of the mxi's.
So again, this is the number
of occurrences of variable xi
in a true or false form.
So I take the maximum occurrence
of any variable, add 1.
That's my base.
It just has to be large enough.
And this is basically the
entire reduction, is one line.
If I have three triples--
if I have a triple xi, xj,
xk, I'm going to convert
that into a number that
looks like this where the one
positions are-- I don't really
know the order, but
they are i, j, and k.
Everything else is zero.
And this is in
base b, not base 2.
It's a little weird.
All my digits are 0 or
1, but I'm in base b.
And three of the digits are 1.
And the rest are zero.
Why?
Because of my target sum.
Target sum is going
to be 1111111111.
So this number,
in algebra, you're
write this as b to the i plus
b to the j plus b to the k.
This you would write as the
sum of b to the i for all i.
Do you see why this works?
It's actually really simple.
For this instance,
my goal is to choose
a subset of these numbers
that add up to this number.
How could that possibly happen?
Well, I've got to choose--
every time I choose
one of the numbers, those three
digits get set to 1 in my sum.
If I ever have a collision,
if I add two 1s together,
I'm going to get a 2.
That's not good,
because once I get a 2,
I'll never be able
to get back to a 1,
because my base is really big.
This base is designed
so that the total-- this
is the total number
of colliding 1s.
So we set it one
larger than that,
which means you'll never get
a carry when you're adding up
in this base.
That's why I set the base to
be something large, not base 2.
Base 2 might work, but
this is much safer.
So what that means is for each
of these 1s in the target sum,
I've got to find a
triple that has those 1s.
And those triples can't overlap.
So that means choosing a set
of numbers that add up to this
is exactly the same as
choosing a set of triples that
covers all of the elements.
Done, super easy once you
have the right problem.
OK, good.
Now why do I call
this weekly NP-hard?
Because these numbers are giant.
If I have n elements
in X, Y, Z over there--
I guess here they're
called xi, yk, zk.
Sorry, maybe I should've
called them that here.
Doesn't matter.
If I have n of those elements
in X union Y union Z,
the number of digits here is n.
So the number of
digits in order n.
This is fine from an
NP-completeness standpoint.
This is polynomial size.
The number of digits in my
numbers is a polynomial.
And this base is
also pretty small.
So if you wrote
it out in binary,
it would also be polynomial.
So just lost a log factor.
But the size of the numbers, the
actual values of the numbers,
is exponential.
With weak NP-hardness,
that's allowed.
With strong NP-hardness,
that's forbidden.
In strong NP-hardness, you
want the values of the numbers
to be polynomial.
So in this case, the
number of bits is small,
but the actual values
are giant, because you
have to exponentiate.
It would be cool.
And this problem is
only weakly NP-hard.
Maybe you actually know
a pseudo-polynomial time
algorithm for this.
It's basically a knapsack.
If these numbers have polynomial
value, then you can basically,
in your subproblems in
dynamic programming,
you can write down
the number t and just
solve it for all values of t.
And it's easy to solve
it in polynomial time,
polynomial in the
integer values.
So we call that
pseudo-polynomial, because it's
not really polynomial.
It's not polynomial in
the number of digits
that you have to
write down the number.
It's Polynomial in the values.
Weak NP-hardness goes together
with pseudo-polynomial.
That's kind of a matching
result. Say look,
pseudo-polynomial is
the best you can do.
You can't hope for a
polynomial because if you
let the numbers get huge, then
the problem is NP-complete.
But if you force the numbers to
be small, this problem is easy.
So subset sum is a little
funny in that sense.
Cool.
Let me tell you about
another problem, partition.
So partition is pretty
much the same set up.
I'm given n integers.
Let's say they're positive.
And I want to know, is
there a subset-- I'm not
given a target sum t.
Target sum is basically forced.
What I would like is the
sum of all the values in S
to equal the sum
of all the values
not in S. That's A minus
S, which in other words
is going to be the sum of
all values in A divided by 2.
So this is called partition
because you're taking a set,
you're splitting it into
two halves of equal sum.
Every element has to go
in one of the two halves.
And they're called S and A minus
S, like cuts in the flow stuff.
And you want those
two halves to have
exactly the same
sum, which means they
will be the sum divided by 2.
So that better be
even, otherwise
it's not going to be possible.
So again, you want to
decide whether this
is possible or
impossible, yes or no.
I claim this problem is
also weakly NP-complete,
and we can reduce from
subset sum to partition.
This is a little interesting
because partition is actually
a special case of subset sum.
It is the case
where t equals this.
Subset sum, you're trying to
solve it no matter what t is.
t is a given input.
So there's more
instances over here.
Some of them, some
of these instances
are the case where t
equals the sum over 2.
Those are partition instances.
So this is like a subset
of the possible inputs
as over there, which
means this problem is
easier than this one--
no harder anyway.
In other words, I can reduce
partition to subset sum.
I just compute this
value and set that to t,
and then leave the a's alone.
That will reduce
partition to subset sum.
But that's not the
direction I want.
I want to reduce from subset
sum, a problem I can prove
is NP-complete, to
partition, because I
want to prove that
partition is NP-complete.
So in this case, there's an easy
reduction in both directions.
This direction is
a little harder.
So reduction from subset sum.
So I'm given a bunch of ai's.
I'm not going to touch them.
And I'm given a target sum t.
And I basically want to make
that target sum into this half.
To do that, I'm going to
add two numbers to my set.
So I'm going to let sigma
be the sum of the given a's.
And then I'm going to add--
so I'm given a1 through an.
I'm going to add an plus 1,
is going to be sigma plus t.
And I'm going to add an plus
2 to be 2 sigma minus t.
Why?
So these are two
basically huge numbers.
Because sigma is
bigger than-- I mean,
it's the sum of all
the numbers, so it's
bigger than all of them.
And so imagine for
a moment that I
put these two in the same
side of the partition.
I put them both in S
or I put them both out
of S. Their sum by
themselves is 3 sigma.
The t's cancel.
Whereas all the other
items, their sum is sigma.
So I'm host.
If I have 3 sigma on one
side and sigma on the other,
I'm not going to
make them equal.
So in fact, these two elements
have to be on opposite sides.
So there's a side
that has sigma plus t.
There's a side has
2 sigma minus t.
And then there's all the other
n items, and some of them
are going to go to
this side, some of them
are going to go to this side.
Their total value is sigma.
Right now this is
close to sigma.
This is close to 2 sigma.
So they have to kind
of meet in the middle.
In fact, what you'll have to do
is add sigma minus t over here
and add t over here.
Think about it for a second.
If I add sigma minus t,
this comes out to 2 sigma.
If I add t to this, this
comes out to 2 sigma.
That would be good
because they're equal.
And notice that this
is sigma minus t.
This is t.
Their sum is sigma.
So in fact, it has
to be like this.
You add something over here, and
sigma minus something over here
for all the other ai's.
And the something has to be t
in order for these two values
to equalize.
So in order to solve this
slightly larger partition
problem, you have to actually
solve the subset sum problem
because you have to construct
a subset that adds up to t.
t was an arbitrary given value.
So this is pretty nifty.
We're adding some values so that
the new target sum is the 50/50
split when we're
given some values that
have an arbitrary target sum.
So partition is
weakly NP-complete.
Let me go to rectangle packing.
So rectangle packing-- I'm
going to draw a picture.
I give you a bunch of
rectangles of varying sizes.
And I give you a
target rectangle.
Let's call it T.
These are the Ri's.
I want to put these
rectangles into this picture
without any overlaps.
Each of these rectangles
here corresponds to one
of the rectangles over here.
So I'll tell you that the sum
of the areas of these rectangles
is equal to the area of T.
And the question is, can you
pack those rectangles into
T without any overlaps,
and therefore without any gaps,
because the areas are exactly
the same.
I claim this problem
is weakly NP-hard-- I
guess NP-complete by
reduction from partition.
This will be super easy
if you followed what
the definition of partition is.
We're given some integers ai.
And we're going to take each of
them and convert them into a,
let's say, 1 by 3ai rectangle.
Three is to avoid some
rotation we'll see.
And then we're also
given the targets.
Oh no, target sum is given.
Target sum is the sum over 2.
But anyway, we're going to
build our target rectangle
to be-- it's actually
going to be really big.
It's going to be 2 by 3 times t.
So this is that thing.
So this is 3/2 sum of the a's.
OK, that's about it.
In order to pack these
rectangles into here,
because each of them
is at least three long,
you cannot pack them vertically.
They have to be horizontal.
So in fact what your
packing will look like is
they'll be the top half
and the bottom half.
And the top half, the total
length of those rectangles
has to add up to 3/2 sum of A.
Everything was scaled up by 3,
so that's 1/2 of A on
the top and the bottom.
That's a partition.
In order to pack the
rectangles into here,
you have to solve the partition
problem, and vice versa.
Easy.
OK, let me show you one
more thing, jigsaw puzzles.
This is not the jigsaw puzzles
you grew up on, somewhat more
generalized.
So a piece is going to
look something like this.
I drew them
intentionally different.
So on each, you
have a unit square.
Some of the sides can be flat.
Some of them can be tabs.
Some of them can be pockets.
Each tab and pocket has a shape.
And they're not in a perfect
matching with each other.
So there could be
seven of these tabs
and seven of these pockets,
all the same shape.
This is what you might call
ambiguous jigsaw puzzles.
Plus, there is no
image on the piece,
so this is like
hardcore jigsaw puzzles.
This is NP-complete.
And what I'd like to do
is to simulate a rectangle
with a bunch of jigsaw pieces.
So it would look
something like this.
If I have a 1 buy
something rectangle,
I'm going to simulate it
with that same something,
little jigsaw pieces.
And I'm going to make these
shapes only match each other.
And so for every
rectangle, they're
going to have a different shape.
This one will be squares.
At that point I ran out of
shapes I can easily draw,
but you get the idea.
Each rectangle has
a different shape.
And so these have to
match to each other.
You can't mix the
tiles, which means you
have to build this rectangle.
You have to build
this rectangle.
And then if the jigsaw
problem is, can you
fit these into a
given rectangle,
then you get rectangle packing.
But this is not a
valid reduction.
You can't reduce from partition.
Why?
Because these numbers are huge.
Remember, the values of
the numbers in my partition
instance are exponential.
So if I have a value ai and it's
exponential in my problem size,
and I tried to make
ai have little tiles,
that means a number
of jigsaw pieces
will be exponential in n.
That's not good.
That's not allowed.
This is why weak
NP-hardness is annoying.
So instead, we need a
strong NP-hard problem.
This is a problem
that's NP-hard even when
the numbers are polynomial
in value, not just in size.
And it's called 4-partition.
4-partition, you're given
n integers, as usual.
Say set is A. And you want
to split those integers
into n over 4 quadruples
of the same sum.
So this would be the sum of
A divided by n over four.
That's your target sum.
So before we had to
split into two parts that
had the same sum.
That was partition.
Now we have to split
into n over 4 parts.
Each part will have exactly
four numbers, four integers.
And they should all
have the same sum.
This problem is hard
even when the integers
have polynomial value.
So the values are at most
some polynomial in n.
I won't prove it here, but
it's in my lecture notes
if you're curious.
It's like this
proof, but harder.
You end up, instead of
having n digit numbers,
you have five digit numbers.
Each digit only has a polynomial
in n different values.
So the total value of the
numbers is only polynomial.
It's like n to the
fifth or something.
Good news is that
this reduction I just
gave you is also a
reduction from 4-partition
because it's the same set up.
Again, I'm given integers.
Each integer I'm going to
represent by that many tiles.
Now the number of tiles
is only polynomial,
so this is a valid reduction.
And again, if I have to
pack all of these tiles
into a rectangular board,
that's exactly the same
as packing these integers.
Well, I guess I should do
rectangle packing again.
So this is a proof rectangle
packing was weakly NP-hard.
But in fact it's
strongly NP-hard.
You just change
these dimensions.
You say well, I need whatever,
n over 4 different parts, each
of size the sum over n over 4.
You need some scale factor here.
Three doesn't work.
Use n or something-- n and n.
That will prove that
rectangle packing is actually
strongly NP-hard because
we're reducing for 4-partition
instead of partition.
And then you can reduce
rectangle packing
to jigsaw puzzles because you
have strong hardness over here.
Over here we don't have numbers.
We just have these pieces.
So whenever you convert
from a number problem
to a non-number problem, if
you're representing the numbers
in unary, which is
what's going on here,
you need strong
NP-hardness for it to work.
Weak NP-hardness isn't enough.
Then we get jigsaw puzzles,
which we know and love,
are NP-complete.
That's it.
