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ERIC DEMAINE: All right,
today we do NP completeness,

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00:00:32,340 --> 00:00:35,420
an entire field in one lecture.

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00:00:35,420 --> 00:00:36,010
Should be fun.

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00:00:36,010 --> 00:00:38,590
I actually taught an entire
class about this topic

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00:00:38,590 --> 00:00:42,620
last semester, but now we're
going to do it in 80 minutes.

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00:00:42,620 --> 00:00:45,610
And we're going to look at
lots of different problems,

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00:00:45,610 --> 00:00:48,440
from Super Mario Brothers
to jigsaw puzzles,

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00:00:48,440 --> 00:00:51,000
and show that
they're NP -complete.

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00:00:51,000 --> 00:00:51,930
This is a fun area.

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00:00:51,930 --> 00:00:54,499
As Srini mentioned last class,
it's all about reductions.

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00:00:54,499 --> 00:00:56,040
It's all about
converting one problem

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00:00:56,040 --> 00:00:59,380
into another, which is a fun
kind of puzzle in itself.

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00:00:59,380 --> 00:01:01,170
It's an algorithmic challenge.

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00:01:01,170 --> 00:01:03,410
And we're going to do it a lot.

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00:01:03,410 --> 00:01:06,600
But first I'm going to remind
you of some of the things you

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00:01:06,600 --> 00:01:10,010
learned from 006, and tell you
what we need to do in order

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00:01:10,010 --> 00:01:14,830
to prove all of these relations,
what exactly we need to show

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00:01:14,830 --> 00:01:18,750
for each of those arrows,
and why it's interesting.

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00:01:18,750 --> 00:01:23,490
So this is generally around
the P versus NP problem.

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00:01:23,490 --> 00:01:26,630
So remember, P is all
the problems we know how

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00:01:26,630 --> 00:01:27,932
to solve in polynomial time.

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00:01:27,932 --> 00:01:30,140
Well not just the ones we
know how to solve, but also

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00:01:30,140 --> 00:01:36,430
the ones that can
be solved, which

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00:01:36,430 --> 00:01:44,490
is pretty much-- which is the
topic of 6.006, and 6.046 up

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00:01:44,490 --> 00:01:45,250
till now.

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00:01:45,250 --> 00:01:47,210
But for now, in the
next few lectures,

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00:01:47,210 --> 00:01:49,480
we'll be talking about
problems that are probably not

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00:01:49,480 --> 00:01:52,940
polynomially solvable,
and what to do about them.

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00:01:52,940 --> 00:01:57,590
Polynomial, as you now, is
like n to the some constant.

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00:01:57,590 --> 00:01:59,350
Polynomial good exponential bad.

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00:02:03,490 --> 00:02:04,300
What is n?

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00:02:04,300 --> 00:02:08,630
I guess n is the size
of the problem, which

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00:02:08,630 --> 00:02:11,150
we'll have to be a little
bit careful about today.

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00:02:11,150 --> 00:02:17,240
And then NP is not
problem solvable

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00:02:17,240 --> 00:02:19,480
not in polynomial time,
but it's problem solvable

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00:02:19,480 --> 00:02:21,800
in nondeterministic
polynomial time.

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00:02:24,720 --> 00:02:27,060
And in this case
we need to focus

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00:02:27,060 --> 00:02:32,010
on a particular type of problem,
which is decision problems.

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00:02:32,010 --> 00:02:35,520
Decision just means that the
answer is either yes or no.

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00:02:35,520 --> 00:02:37,450
So it's a single bit answer.

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00:02:43,130 --> 00:02:44,935
We will see why we
need to restrict

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00:02:44,935 --> 00:02:46,435
to that kind of
problem in a moment.

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00:03:11,030 --> 00:03:14,370
So this is problems you can
solve in polynomial time.

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00:03:14,370 --> 00:03:16,960
Same notion of polynomials,
same notion of n,

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00:03:16,960 --> 00:03:19,640
but in a totally unrealistic
model of computation.

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00:03:19,640 --> 00:03:22,080
Which is a
nondeterministic model.

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00:03:22,080 --> 00:03:24,060
In a nondeterministic
model, what

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00:03:24,060 --> 00:03:29,150
you can do is say
instead of computing

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00:03:29,150 --> 00:03:33,010
something from something you
know, you could make a guess.

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00:03:33,010 --> 00:03:58,840
So you can guess one out of
polynomially many options

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00:03:58,840 --> 00:04:02,090
in constant time.

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00:04:02,090 --> 00:04:04,680
So normally a constant time
operation, in regular models,

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00:04:04,680 --> 00:04:08,230
like you add two numbers, or you
do an if, that sort of thing.

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00:04:08,230 --> 00:04:10,870
Here we can make a guess.

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00:04:10,870 --> 00:04:15,290
I give the computer polynomially
many options I'm interested in.

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00:04:15,290 --> 00:04:17,050
Computer's going to
give me one of them.

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00:04:17,050 --> 00:04:20,570
It's going to give
me a good guess.

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00:04:20,570 --> 00:04:21,959
Guess is guaranteed to be good.

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00:04:21,959 --> 00:04:24,290
And good means here
that I want to get

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00:04:24,290 --> 00:04:27,110
to a yes answer if I can.

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00:04:27,110 --> 00:04:35,630
So the formal statement
is, if any guess

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00:04:35,630 --> 00:04:46,960
would lead to a yes answer,
then we get such a guess.

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00:04:51,096 --> 00:04:52,580
OK, this is weird.

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00:04:52,580 --> 00:04:53,610
And it's asymmetric.

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00:04:53,610 --> 00:04:55,060
It's biased towards yes.

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00:04:55,060 --> 00:04:58,010
And this is why we can only
think about decision problems,

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00:04:58,010 --> 00:04:59,090
yes or no.

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00:04:59,090 --> 00:05:00,360
You could bias towards no.

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00:05:00,360 --> 00:05:02,060
You get something
else called coNP.

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00:05:02,060 --> 00:05:04,560
But we'll focus here just on NP.

78
00:05:04,560 --> 00:05:08,820
So the idea is I'd really
like to find a guess that

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00:05:08,820 --> 00:05:09,980
leads to a yes answer.

80
00:05:09,980 --> 00:05:12,670
And the machine magically
gives me one if there is one.

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00:05:12,670 --> 00:05:15,170
Which means if I end
up saying no, that

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00:05:15,170 --> 00:05:18,790
means there was absolutely no
path that would lead to a yes.

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00:05:18,790 --> 00:05:21,300
So when you get a no, you
get a lot of information.

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00:05:21,300 --> 00:05:23,250
When you get a yes, you
get some information.

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00:05:23,250 --> 00:05:24,249
But hey, you were lucky.

86
00:05:24,249 --> 00:05:25,700
Hard to complain.

87
00:05:25,700 --> 00:05:30,340
So in 006, I often call this
the lucky model of computation.

88
00:05:30,340 --> 00:05:31,960
That's the informal version.

89
00:05:31,960 --> 00:05:35,740
But nondeterminism is
what's really going on here.

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00:05:35,740 --> 00:05:41,820
So maybe it's useful
to get an example.

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00:05:41,820 --> 00:05:47,240
So here's a problem we'll--
this is sort of the granddaddy

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00:05:47,240 --> 00:05:49,080
of all NP-complete problems.

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00:05:49,080 --> 00:05:51,560
We'll get to
completeness in a moment.

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00:05:51,560 --> 00:06:01,855
3SAT-- SAT stands
for satisfiability.

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00:06:10,030 --> 00:06:12,590
So in 3SAT, the
input to the problem

96
00:06:12,590 --> 00:06:14,470
looks something like this.

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00:06:14,470 --> 00:06:16,016
I'm just going to
give an example.

98
00:06:26,930 --> 00:06:30,870
And in case you've forgotten
your weird logic notation,

99
00:06:30,870 --> 00:06:33,520
this is an and.

100
00:06:33,520 --> 00:06:36,580
These are ORs.

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00:06:36,580 --> 00:06:42,680
And I'm using this
for negation, not.

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00:06:42,680 --> 00:06:48,050
So in other words, I'm given
a formula which is and of ORs.

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00:06:48,050 --> 00:06:52,340
And each or clause only
has three things in it.

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00:06:52,340 --> 00:06:53,715
These things are
called literals.

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00:07:00,300 --> 00:07:03,990
And a literal is either
a variable x sub i,

106
00:07:03,990 --> 00:07:08,020
or it's the negation of
a variable, not x sub i.

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00:07:08,020 --> 00:07:09,500
So this is a typical example.

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00:07:09,500 --> 00:07:11,380
You could have no negations.

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00:07:11,380 --> 00:07:13,510
You could here have one
negation, two negations,

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00:07:13,510 --> 00:07:15,950
any number of
negations per clause.

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00:07:15,950 --> 00:07:19,580
These groups of three--
these or of three

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00:07:19,580 --> 00:07:23,900
things, three literals,
are called clauses.

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00:07:23,900 --> 00:07:26,960
And they're all ANDed together.

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00:07:26,960 --> 00:07:30,100
And my goal is, this should
be a decision question,

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00:07:30,100 --> 00:07:31,960
so I have a yes or no question.

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00:07:31,960 --> 00:07:44,030
And that question is, can
you set the variables--

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00:07:44,030 --> 00:07:52,010
So they're x1 to true or false?

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00:07:52,010 --> 00:07:56,020
So each variable I get to choose
a true or false designation

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00:07:56,020 --> 00:07:57,860
such that the formula
comes out true.

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00:08:04,870 --> 00:08:07,390
I use T and F for
true and false.

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00:08:07,390 --> 00:08:09,650
So I want to set
these variables such

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00:08:09,650 --> 00:08:12,230
that every clause comes
out true, because they're

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00:08:12,230 --> 00:08:12,970
ANDed together.

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00:08:12,970 --> 00:08:15,910
So I have to satisfy this
clause in one of three ways.

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00:08:15,910 --> 00:08:17,700
Maybe I satisfy
it all three ways.

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00:08:17,700 --> 00:08:19,700
Doesn't matter, as long
as at least one of these

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00:08:19,700 --> 00:08:22,490
should be true, and at least
one of these should be true,

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00:08:22,490 --> 00:08:25,940
and at least one of each
clause should be true.

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00:08:25,940 --> 00:08:28,300
So that's the 3SAT problem.

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00:08:28,300 --> 00:08:29,421
This is a hard problem.

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00:08:29,421 --> 00:08:31,170
We don't know a
polynomial time algorithm.

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00:08:31,170 --> 00:08:32,520
There probably isn't one.

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00:08:32,520 --> 00:08:37,669
But there is a polynomial time
nondeterministic algorithm.

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00:08:37,669 --> 00:08:54,140
So this problem is in NP
because if I have lucky guesses,

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00:08:54,140 --> 00:08:56,460
it's kind of designed to
solve this kind of problem.

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00:08:56,460 --> 00:09:03,730
What I'm going to do is guess
whether x1 is true or false.

137
00:09:03,730 --> 00:09:05,490
So I have two choices.

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00:09:05,490 --> 00:09:08,240
And I'm going to ask my machine
to make the right choice,

139
00:09:08,240 --> 00:09:10,120
whether it should
be true or false.

140
00:09:10,120 --> 00:09:13,760
Then I'll guess x2.

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00:09:13,760 --> 00:09:17,480
Each of these guess operations
takes constant time.

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00:09:17,480 --> 00:09:19,250
So I do it for every variable.

143
00:09:19,250 --> 00:09:21,000
And then I'm going
to check whether I

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00:09:21,000 --> 00:09:22,670
happen to satisfy the formula.

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00:09:28,960 --> 00:09:33,140
And if it comes out true,
then I'll return yes.

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00:09:33,140 --> 00:09:35,267
And if it comes out
false, I'll return no.

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00:09:43,720 --> 00:09:47,720
And because NP is biased
towards yes answers,

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00:09:47,720 --> 00:09:50,824
it always finds a yes
answer if you can.

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00:09:50,824 --> 00:09:53,560
If there's some way to
satisfy the formula,

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00:09:53,560 --> 00:09:55,160
then I will get it.

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00:09:55,160 --> 00:09:57,550
If there's some way to make
the formula come out true,

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00:09:57,550 --> 00:09:59,890
then this algorithm
will return yes.

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00:09:59,890 --> 00:10:03,930
If there's no way
to satisfy it, then

154
00:10:03,930 --> 00:10:05,996
this nondeterministic
algorithm will return no.

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00:10:05,996 --> 00:10:07,370
That's just the
definition of how

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00:10:07,370 --> 00:10:09,340
nondeterministic machines work.

157
00:10:09,340 --> 00:10:10,410
It's a little weird.

158
00:10:10,410 --> 00:10:13,670
But you can see from
this kind of prototype

159
00:10:13,670 --> 00:10:17,800
of a nondeterministic algorithm,
you can actually always

160
00:10:17,800 --> 00:10:20,920
arrange for your guessing
to be at the beginning.

161
00:10:20,920 --> 00:10:25,830
And then you do some regular
polynomial time checking

162
00:10:25,830 --> 00:10:29,010
or deterministic checking.

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00:10:29,010 --> 00:10:30,440
So when you rewrite
your algorithm

164
00:10:30,440 --> 00:10:32,820
like this with guesses up
front and then checking,

165
00:10:32,820 --> 00:10:36,360
you can also think of it as
a verification algorithm.

166
00:10:36,360 --> 00:10:41,050
So you can say, your friend
claims that this 3SAT formula

167
00:10:41,050 --> 00:10:43,420
is satisfiable,
meaning there's a way

168
00:10:43,420 --> 00:10:46,140
to set the variable so
that it comes out true.

169
00:10:46,140 --> 00:10:48,264
So this is called a
satisfying assignment.

170
00:10:50,990 --> 00:10:52,627
Satisfying just means make true.

171
00:10:56,830 --> 00:10:59,355
And you're like, no,
I don't believe you.

172
00:10:59,355 --> 00:11:01,940
And your friend says no,
no, no, really, it's true.

173
00:11:01,940 --> 00:11:03,460
And here's how I can prove it.

174
00:11:03,460 --> 00:11:04,720
You set x1 to false.

175
00:11:04,720 --> 00:11:05,725
You set x2 to true.

176
00:11:05,725 --> 00:11:09,260
You set x3-- basically
they give you the guesses.

177
00:11:09,260 --> 00:11:11,605
And then you don't
have to be convinced

178
00:11:11,605 --> 00:11:12,980
that those are
the right guesses,

179
00:11:12,980 --> 00:11:14,646
you can check that
it's the right guess.

180
00:11:14,646 --> 00:11:17,090
You can compute this
formula in linear time,

181
00:11:17,090 --> 00:11:18,510
see what the outcome is.

182
00:11:18,510 --> 00:11:20,420
If someone tells you
what the xi's are,

183
00:11:20,420 --> 00:11:21,980
you can very quickly
see whether that

184
00:11:21,980 --> 00:11:24,310
was a satisfying assignment.

185
00:11:24,310 --> 00:11:26,440
So you could call
this a solution,

186
00:11:26,440 --> 00:11:28,650
and then there's a
polynomial time verification

187
00:11:28,650 --> 00:11:32,700
algorithm that checks
that solutions are valid.

188
00:11:32,700 --> 00:11:36,170
But, you can only do
that for yes answers.

189
00:11:36,170 --> 00:11:39,510
Your friend says no,
this is not satisfiable,

190
00:11:39,510 --> 00:11:42,270
they have no way of
proving it to you.

191
00:11:42,270 --> 00:11:46,450
I mean, other than checking
all the assignments separately,

192
00:11:46,450 --> 00:11:48,290
which would take
exponential time,

193
00:11:48,290 --> 00:11:51,140
there's no easy way to confirm
that the answer to this problem

194
00:11:51,140 --> 00:11:51,670
is no.

195
00:11:51,670 --> 00:11:53,470
But there is an
easy way to check

196
00:11:53,470 --> 00:11:55,830
that the answer is yes, namely
I give you the satisfying

197
00:11:55,830 --> 00:11:57,600
assignment.

198
00:11:57,600 --> 00:12:01,130
So this definition of NP
is what I'll stick to.

199
00:12:01,130 --> 00:12:05,100
It's this sort of-- I
like guessing because it's

200
00:12:05,100 --> 00:12:06,500
like dynamic programming.

201
00:12:06,500 --> 00:12:09,329
With dynamic programming
we also guess,

202
00:12:09,329 --> 00:12:11,620
and guessing actually originally
comes from this world,

203
00:12:11,620 --> 00:12:13,390
nondeterminism.

204
00:12:13,390 --> 00:12:16,790
In dynamic programming, we
don't allow this kind of model.

205
00:12:16,790 --> 00:12:19,030
And so we have to check
the guesses separately.

206
00:12:19,030 --> 00:12:20,790
And so we spend lots of time.

207
00:12:20,790 --> 00:12:22,680
Here, magically, you
always get the right

208
00:12:22,680 --> 00:12:24,320
guess in only constant time.

209
00:12:24,320 --> 00:12:26,070
So this is a much
more powerful model.

210
00:12:26,070 --> 00:12:30,251
Of course there's no computers
that work like this, sadly,

211
00:12:30,251 --> 00:12:32,280
or I guess more interestingly.

212
00:12:32,280 --> 00:12:36,870
So this is more about confirming
that your problem is not

213
00:12:36,870 --> 00:12:38,830
totally impossible.

214
00:12:38,830 --> 00:12:42,560
At least you can check the
answers in polynomial time.

215
00:12:42,560 --> 00:12:46,830
So that's one thing.

216
00:13:31,110 --> 00:13:33,210
So this is an equivalent
definition of NP

217
00:13:33,210 --> 00:13:35,430
because you can take a
nondeterministic algorithm

218
00:13:35,430 --> 00:13:37,300
and put the guessing up top.

219
00:13:37,300 --> 00:13:39,200
You can call the
results of those guesses

220
00:13:39,200 --> 00:13:42,060
a certificate that
an answer is yes.

221
00:13:42,060 --> 00:13:45,940
And then you have a regular old
deterministic polynomial time

222
00:13:45,940 --> 00:13:48,430
algorithm that, given
that certificate,

223
00:13:48,430 --> 00:13:54,516
will verify that it actually
proves that the answer is yes.

224
00:13:57,820 --> 00:14:00,730
It's just that certificate
has to be polynomial size.

225
00:14:00,730 --> 00:14:02,960
You can't guess something
of exponential size.

226
00:14:02,960 --> 00:14:07,010
You can only guess something of
polynomial size in this model.

227
00:14:07,010 --> 00:14:10,410
So seems a little weird.

228
00:14:10,410 --> 00:14:16,610
But we'll see why this is
useful in a little bit.

229
00:14:16,610 --> 00:14:18,795
So let me go to NP completeness.

230
00:14:24,740 --> 00:14:37,320
So if I have a problem X,
it's NP-complete if X is in NP

231
00:14:37,320 --> 00:14:38,830
and X is NP-hard.

232
00:14:41,760 --> 00:14:44,130
But I haven't told
you what NP-hard is.

233
00:14:44,130 --> 00:15:24,860
Maybe you remember from
006, but let me remind you.

234
00:15:24,860 --> 00:15:28,840
So, I need to define reduce.

235
00:15:28,840 --> 00:15:30,684
So maybe I'll do
that as well, then

236
00:15:30,684 --> 00:15:31,850
we can talk about all these.

237
00:16:26,130 --> 00:16:29,110
OK, a lot of definitions.

238
00:16:29,110 --> 00:16:32,970
But the idea of NP
hardness is very simple.

239
00:16:32,970 --> 00:16:35,140
If problem X is
NP-hard, it means

240
00:16:35,140 --> 00:16:39,890
that it's at least
as hard as-- sorry,

241
00:16:39,890 --> 00:16:46,280
that is a Y-- it's at least
as hard as all problems in NP.

242
00:16:46,280 --> 00:16:48,140
Intuitively, X means
it's at least as

243
00:16:48,140 --> 00:16:50,280
hard as everything in NP.

244
00:16:50,280 --> 00:16:53,310
Whereas being in NP is
a positive statement.

245
00:16:53,310 --> 00:16:55,240
That says it's not
too hard, at least

246
00:16:55,240 --> 00:16:57,760
there's a polynomial time
verification algorithm.

247
00:16:57,760 --> 00:17:00,060
So being in NP is good news.

248
00:17:00,060 --> 00:17:02,650
It says you're no
harder than NP.

249
00:17:02,650 --> 00:17:05,920
NP-hard says you're at least
as hard as everything in NP.

250
00:17:05,920 --> 00:17:08,230
And so NP-complete
is a nice answer

251
00:17:08,230 --> 00:17:11,579
because this says you're exactly
as hard as everything in NP--

252
00:17:11,579 --> 00:17:13,849
no harder, no easier.

253
00:17:13,849 --> 00:17:17,099
If you draw, in
this vague sense,

254
00:17:17,099 --> 00:17:23,589
computational difficulty
on one axis-- which is not

255
00:17:23,589 --> 00:17:26,380
really accurate, but I
like to do it anyway--

256
00:17:26,380 --> 00:17:31,320
and you have P is all of
these easy problems down here.

257
00:17:31,320 --> 00:17:35,460
And NP is some
larger set like this.

258
00:17:35,460 --> 00:17:38,139
NP-hard is from here over.

259
00:17:42,540 --> 00:17:45,040
And this point right
here is NP-complete.

260
00:17:49,500 --> 00:17:53,350
Being in NP means you're left
of this line, or on the line.

261
00:17:53,350 --> 00:17:55,600
And being NP-hard means
you're right of this line,

262
00:17:55,600 --> 00:17:56,480
or on the line.

263
00:17:56,480 --> 00:17:58,610
NP-complete means
you're right there.

264
00:17:58,610 --> 00:18:02,510
So that's a very definitive
sense of hardness.

265
00:18:02,510 --> 00:18:04,690
Now there is this
slight catch, which

266
00:18:04,690 --> 00:18:06,550
is we don't know
whether P equals NP.

267
00:18:06,550 --> 00:18:11,450
So maybe this is the same
as this, but probably not.

268
00:18:11,450 --> 00:18:14,530
Unless you believe
in luck, basically,

269
00:18:14,530 --> 00:18:16,400
unless you imagine
that a computer could

270
00:18:16,400 --> 00:18:19,830
engineer luck and always
guess the right things

271
00:18:19,830 --> 00:18:23,630
without spending a lot of
time, then P does not equal NP.

272
00:18:23,630 --> 00:18:26,680
And in that world, what
we get is that if you have

273
00:18:26,680 --> 00:18:29,650
an NP-complete problem, or
actually any NP-hard problem,

274
00:18:29,650 --> 00:18:33,590
you know it cannot be NP.

275
00:18:33,590 --> 00:18:38,240
So if you have
that X is NP-hard,

276
00:18:38,240 --> 00:18:44,405
then you know that X is
not in P unless all of NP

277
00:18:44,405 --> 00:18:47,690
is in P. So unless P equals NP.

278
00:18:47,690 --> 00:18:51,040
And most reasonable people
do not believe this.

279
00:18:51,040 --> 00:18:53,160
And so instead they
have to believe this,

280
00:18:53,160 --> 00:18:57,580
that your problem is not
polynomially solvable.

281
00:18:57,580 --> 00:18:59,740
So why is this true?

282
00:18:59,740 --> 00:19:01,680
Because if your
problem is NP-hard,

283
00:19:01,680 --> 00:19:05,880
it is at least as hard
as every problem in NP.

284
00:19:05,880 --> 00:19:08,449
And if you believe that
there is some problem in NP--

285
00:19:08,449 --> 00:19:09,990
we don't necessarily
know which one--

286
00:19:09,990 --> 00:19:14,280
but if there is any problem out
there in NP that is not in P,

287
00:19:14,280 --> 00:19:18,710
then X has to be at
least as hard as it.

288
00:19:18,710 --> 00:19:22,730
So it also requires
nonpolynomial time, something

289
00:19:22,730 --> 00:19:25,060
larger than polynomial time.

290
00:19:25,060 --> 00:19:27,090
What does at least
as hard mean though?

291
00:19:27,090 --> 00:19:29,690
We're going to define it
in terms of reductions.

292
00:19:29,690 --> 00:19:31,530
Reduction from one
problem to another

293
00:19:31,530 --> 00:19:33,910
is just a polynomial
time algorithm,

294
00:19:33,910 --> 00:19:36,130
regular deterministic
polynomial time,

295
00:19:36,130 --> 00:19:38,440
that converts an
input to the problem A

296
00:19:38,440 --> 00:19:40,980
into an equivalent
input to problem

297
00:19:40,980 --> 00:19:45,554
B. Equivalent means that it
has the same yes or no answer.

298
00:19:48,105 --> 00:19:50,480
And we'll just be thinking
about decision problems today.

299
00:19:55,410 --> 00:19:59,610
So why would I care
about a reduction?

300
00:19:59,610 --> 00:20:03,960
Because what it tells me is that
if I know how to solve problem

301
00:20:03,960 --> 00:20:07,860
B, then I also know how
to solve problem A. If I

302
00:20:07,860 --> 00:20:11,180
have a, say, a polynomial
time algorithm for solving B

303
00:20:11,180 --> 00:20:13,960
and I want one for A,
I just take my A input.

304
00:20:13,960 --> 00:20:16,440
I convert it into the
equivalent B input.

305
00:20:16,440 --> 00:20:18,190
Then I run my algorithm
for B, and then it

306
00:20:18,190 --> 00:20:19,880
gives me the answer
to the A problem

307
00:20:19,880 --> 00:20:22,920
because the answers
are the same.

308
00:20:22,920 --> 00:20:27,480
So if you have a reduction
like this and if say,

309
00:20:27,480 --> 00:20:29,940
B, has a polynomial
time algorithm,

310
00:20:29,940 --> 00:20:35,410
then so does A, because you
can just convert A into B,

311
00:20:35,410 --> 00:20:39,870
and then solve B.
Also this works

312
00:20:39,870 --> 00:20:42,990
for nondeterministic algorithms.

313
00:20:48,540 --> 00:20:50,288
Not too important.

314
00:20:53,360 --> 00:20:56,710
So what this tells us is
that in a certain sense--

315
00:20:56,710 --> 00:21:00,850
get this right--
well this is saying,

316
00:21:00,850 --> 00:21:06,320
if I can solve B,
that I can solve A.

317
00:21:06,320 --> 00:21:12,020
So this is saying
that B is at least as

318
00:21:12,020 --> 00:21:18,020
hard as A. I think I got
that right, a little tricky.

319
00:21:18,020 --> 00:21:21,210
So if we want to prove
the problem is NP hard,

320
00:21:21,210 --> 00:21:24,560
what we do is show that
every problem in NP

321
00:21:24,560 --> 00:21:28,150
can be reduced to the problem
of X. So now we can go back

322
00:21:28,150 --> 00:21:31,390
and say well, if we believe
that there is some problem Y,

323
00:21:31,390 --> 00:21:34,520
that is in NP minus P, if
there's something out here that

324
00:21:34,520 --> 00:21:38,780
is not in P, then we
can take that problem Y,

325
00:21:38,780 --> 00:21:40,790
and by this definition,
we can reduce it

326
00:21:40,790 --> 00:21:43,750
to X, because everything
in NP reduces to X.

327
00:21:43,750 --> 00:21:49,250
And so then I can
solve my problem

328
00:21:49,250 --> 00:21:51,420
Y, which is in NP minus P,

329
00:21:51,420 --> 00:21:52,810
by converting it
to X and solving

330
00:21:52,810 --> 00:21:55,770
X. So that means X better
not have a polynomial time

331
00:21:55,770 --> 00:21:59,240
algorithm, because if
it did, Y would also

332
00:21:59,240 --> 00:22:00,650
have a polynomial
time algorithm.

333
00:22:00,650 --> 00:22:02,740
And then in general,
P would equal

334
00:22:02,740 --> 00:22:05,730
NP, because every problem
in NP can be converted to X.

335
00:22:05,730 --> 00:22:07,780
So if X has a polynomial
time algorithm,

336
00:22:07,780 --> 00:22:10,470
then every problem Y does.

337
00:22:10,470 --> 00:22:11,667
Question?

338
00:22:11,667 --> 00:22:13,250
AUDIENCE: For the
second if statement,

339
00:22:13,250 --> 00:22:17,690
why can't you say that if
A is in NP, B is in NP?

340
00:22:17,690 --> 00:22:20,190
ERIC DEMAINE: So you're asked
us about the reverse question.

341
00:22:20,190 --> 00:22:23,900
If is A in NP, can we
conclude that B is in NP?

342
00:22:23,900 --> 00:22:25,314
And the answer is no.

343
00:22:30,890 --> 00:22:33,880
Because this reduction only
lets us convert from A to B.

344
00:22:33,880 --> 00:22:37,100
It doesn't let us do anything
for converting from B to A.

345
00:22:37,100 --> 00:22:39,580
So if we know how to
solve A and we also

346
00:22:39,580 --> 00:22:43,180
know how to convert A into B,
it doesn't tell us anything.

347
00:22:43,180 --> 00:22:47,345
It could be B is a much
harder problem than A,

348
00:22:47,345 --> 00:22:48,450
in that situation.

349
00:22:53,394 --> 00:22:55,310
That's, I think, as good
as I can do for that.

350
00:22:55,310 --> 00:22:55,976
Other questions?

351
00:22:58,891 --> 00:22:59,390
All right.

352
00:22:59,390 --> 00:23:01,860
It is really tricky to get
these directions right.

353
00:23:01,860 --> 00:23:09,410
So let me give you a handy guide
on how to not make a mistake.

354
00:23:09,410 --> 00:23:11,028
So maybe over here.

355
00:23:20,410 --> 00:23:24,430
What we care about, from
an algorithmic perspective,

356
00:23:24,430 --> 00:23:27,780
is proving the problems
are NP-complete.

357
00:23:37,716 --> 00:23:39,590
Because if we prove
NP-completeness-- I mean,

358
00:23:39,590 --> 00:23:41,044
really we care
about NP-hardness,

359
00:23:41,044 --> 00:23:42,710
but we might as well
do NP-completeness.

360
00:23:42,710 --> 00:23:45,250
Most of the problems that
we'll see that are NP-hard

361
00:23:45,250 --> 00:23:47,950
are also NP-complete.

362
00:23:47,950 --> 00:23:51,450
So when we prove this, we
prove that there is basically

363
00:23:51,450 --> 00:23:53,990
no polynomial time
algorithm for that problem.

364
00:23:53,990 --> 00:23:55,810
So that's good to
know, because then we

365
00:23:55,810 --> 00:23:59,840
can just give up searching for
a polynomial time algorithm.

366
00:23:59,840 --> 00:24:02,300
So all the problems
we've seen so far

367
00:24:02,300 --> 00:24:05,050
have polynomial time algorithms,
except a couple in your problem

368
00:24:05,050 --> 00:24:07,089
sets, which were
actually NP-complete.

369
00:24:07,089 --> 00:24:09,130
And the best you could
have done was exponential,

370
00:24:09,130 --> 00:24:11,420
unless P equals NP.

371
00:24:11,420 --> 00:24:15,550
So here's how you can prove
this kind of lower bound

372
00:24:15,550 --> 00:24:17,550
to say look, I don't need
to look for algorithms

373
00:24:17,550 --> 00:24:19,970
any more because my
problem is just too hard.

374
00:24:19,970 --> 00:24:22,580
It's as hard as
everything in NP.

375
00:24:22,580 --> 00:24:25,700
So this is just a summary
of those definitions.

376
00:24:25,700 --> 00:24:29,820
The first thing you do
is prove that X is in NP.

377
00:24:29,820 --> 00:24:33,140
The second thing you do is
prove that X is NP-hard.

378
00:24:33,140 --> 00:24:39,460
And to do that, you reduce
from some known NP-complete

379
00:24:39,460 --> 00:24:44,430
problem-- or I guess
NP-hard, but we'll

380
00:24:44,430 --> 00:24:55,600
use NP-complete--
to your problem

381
00:24:55,600 --> 00:25:06,030
X. Maybe I'll give
this a name Y.

382
00:25:06,030 --> 00:25:08,080
OK, so to prove
that X is in NP, you

383
00:25:08,080 --> 00:25:09,970
do something like
what we did over here,

384
00:25:09,970 --> 00:25:12,390
which is to give a
nondeterministic algorithm.

385
00:25:12,390 --> 00:25:13,970
Or you can think
of it as defining

386
00:25:13,970 --> 00:25:19,290
what the certificate is and
then giving a polynomial time

387
00:25:19,290 --> 00:25:22,050
verification algorithm.

388
00:25:22,050 --> 00:25:24,812
So sort of two approaches.

389
00:25:24,812 --> 00:25:29,710
You can give a nondeterministic
polynomial time algorithm,

390
00:25:29,710 --> 00:25:33,234
or you give a certificate
and a verifier.

391
00:25:39,360 --> 00:25:41,230
There's no right or
wrong certificate.

392
00:25:41,230 --> 00:25:43,730
I mean, a certificate, you
can define however you want,

393
00:25:43,730 --> 00:25:46,190
as long as the verifier
can actually check it

394
00:25:46,190 --> 00:25:49,480
and when it says yes, then the
answer to the problem was yes.

395
00:25:49,480 --> 00:25:51,900
So it's really the same thing.

396
00:25:51,900 --> 00:25:53,710
Just want to say
there's some certificate

397
00:25:53,710 --> 00:25:57,236
that a verifier
could actually check.

398
00:25:57,236 --> 00:25:59,110
So that's proving that
your problem is in NP.

399
00:25:59,110 --> 00:26:01,660
It's sort of an
algorithmic thing.

400
00:26:01,660 --> 00:26:03,490
The second part is
all about reductions.

401
00:26:03,490 --> 00:26:05,270
Now the definition
says that I should

402
00:26:05,270 --> 00:26:08,776
reduce every problem
in NP to my problem X.

403
00:26:08,776 --> 00:26:10,150
That's tedious,
because there are

404
00:26:10,150 --> 00:26:11,610
a lot of problems in the world.

405
00:26:11,610 --> 00:26:14,530
So I don't want to do it
for every problem in NP.

406
00:26:14,530 --> 00:26:16,030
I'd like to just do it for one.

407
00:26:16,030 --> 00:26:17,994
Now if I reduce
sorting to my problem,

408
00:26:17,994 --> 00:26:19,160
that's not very interesting.

409
00:26:19,160 --> 00:26:22,980
It says my problem is at
least as hard as sorting.

410
00:26:22,980 --> 00:26:24,940
But I already know
how to solve sorting.

411
00:26:24,940 --> 00:26:28,960
But if I start from an
NP-complete problem,

412
00:26:28,960 --> 00:26:32,720
then I know, by the definition,
that every problem in NP

413
00:26:32,720 --> 00:26:34,550
can be reduced to that problem.

414
00:26:34,550 --> 00:26:37,770
And if I show how to reduce
the NP-complete problem to me,

415
00:26:37,770 --> 00:26:39,830
then I know that
I'm NP-complete too.

416
00:26:39,830 --> 00:26:44,180
Because if I have
any problem Z in NP,

417
00:26:44,180 --> 00:26:49,510
by the definition of NP-complete
of Y I can reduce that to Y.

418
00:26:49,510 --> 00:26:53,330
And then if I can build
a reduction from Y to X,

419
00:26:53,330 --> 00:26:55,520
then I get this reduction.

420
00:26:55,520 --> 00:26:58,380
And so that means I can
convert any problem in NP

421
00:26:58,380 --> 00:27:01,310
to my problem X, which
means X is NP-hard.

422
00:27:01,310 --> 00:27:03,190
That's the definition.

423
00:27:03,190 --> 00:27:06,650
So all this is to
say the first time

424
00:27:06,650 --> 00:27:09,740
you prove a problem is
NP-complete in the world-- this

425
00:27:09,740 --> 00:27:14,700
happened in the '70s by Cook.

426
00:27:14,700 --> 00:27:16,900
Basically he proved that
3SAT is NP-complete.

427
00:27:19,725 --> 00:27:21,100
That was annoying,
because he had

428
00:27:21,100 --> 00:27:23,106
to start from any
problem in NP, and he

429
00:27:23,106 --> 00:27:25,230
had to show that you could
reduce any problem in NP

430
00:27:25,230 --> 00:27:26,810
to 3SAT.

431
00:27:26,810 --> 00:27:29,820
But now that that hard work is
done, our life is much easier.

432
00:27:29,820 --> 00:27:31,780
And in this class all
you need to think about

433
00:27:31,780 --> 00:27:34,620
is picking your favorite
NP-complete problem.

434
00:27:34,620 --> 00:27:36,900
3SAT's a good choice
for almost anything,

435
00:27:36,900 --> 00:27:40,190
but we'll see a bunch of other
problems today from here.

436
00:27:40,190 --> 00:27:44,520
And then reduce from that
known problem to your problem

437
00:27:44,520 --> 00:27:46,690
that you're trying
to prove is NP-hard.

438
00:27:46,690 --> 00:27:50,220
If you can do that, you know
your problem is NP-hard.

439
00:27:50,220 --> 00:27:52,970
So we only need one reduction
for each hardness result, which

440
00:27:52,970 --> 00:27:53,470
is nice.

441
00:27:53,470 --> 00:27:56,060
And this picture is a
collection of reductions.

442
00:27:56,060 --> 00:27:57,360
We're going to start from 3SAT.

443
00:27:57,360 --> 00:27:59,010
I'm not going to prove
that it's NP-complete,

444
00:27:59,010 --> 00:28:01,150
although I'll give you a
hint as to why that's true.

445
00:28:01,150 --> 00:28:02,850
We're going to reduce it
to Super Mario Brothers.

446
00:28:02,850 --> 00:28:04,890
We're going to reduce it to
three dimensional matching.

447
00:28:04,890 --> 00:28:06,290
We're going to reduce
three dimensional matching

448
00:28:06,290 --> 00:28:08,490
to subsets sum, to partition,
to rectangle packing,

449
00:28:08,490 --> 00:28:10,165
to jig saw puzzles.

450
00:28:10,165 --> 00:28:13,310
And we're going to do all
those reductions, hopefully.

451
00:28:13,310 --> 00:28:18,920
And that's proving NP-hardness
of all those problems.

452
00:28:18,920 --> 00:28:20,070
They're also all in NP.

453
00:28:24,990 --> 00:28:31,610
So 30 second intuition
why 3SAT is NP-hard.

454
00:28:31,610 --> 00:28:35,310
Well, if you have
any problem in NP,

455
00:28:35,310 --> 00:28:39,580
that means there is one of these
nondeterministic polynomial

456
00:28:39,580 --> 00:28:44,870
time algorithms, or there
is some verifier given

457
00:28:44,870 --> 00:28:46,740
a polynomial size certificate.

458
00:28:46,740 --> 00:28:49,250
So that verifier is
just some algorithm.

459
00:28:49,250 --> 00:28:52,070
And software and hardware
are basically the same thing,

460
00:28:52,070 --> 00:28:52,570
right?

461
00:28:52,570 --> 00:28:54,200
So you can convert
that algorithm

462
00:28:54,200 --> 00:28:57,150
into a circuit that
implements the algorithm.

463
00:28:57,150 --> 00:28:59,840
And if I have a circuit with
like ANDs and ORs and NOTs,

464
00:28:59,840 --> 00:29:02,469
I can convert that
into a Boolean formula

465
00:29:02,469 --> 00:29:03,510
with ANDs, ORs, and NOTs.

466
00:29:03,510 --> 00:29:06,030
Circuits and formulas
are about the same.

467
00:29:06,030 --> 00:29:08,374
And if I have a
formula-- fun fact,

468
00:29:08,374 --> 00:29:10,040
although this is a
little less obvious--

469
00:29:10,040 --> 00:29:17,120
you can convert it into this
form, an AND of triple ORs.

470
00:29:17,120 --> 00:29:18,850
And once you've done
that, that formula

471
00:29:18,850 --> 00:29:22,120
is equivalent to the
original algorithm.

472
00:29:22,120 --> 00:29:25,840
And the inputs to that
verification algorithm,

473
00:29:25,840 --> 00:29:29,220
the certificate, are represented
by these variables, the xi's.

474
00:29:29,220 --> 00:29:31,040
And so deciding whether
there's some way

475
00:29:31,040 --> 00:29:33,160
to set the xi's to
make the formula true

476
00:29:33,160 --> 00:29:37,040
is the same thing as saying is
there some certificate where

477
00:29:37,040 --> 00:29:40,220
the verifier says yes, which
is the same thing as saying

478
00:29:40,220 --> 00:29:44,120
that the problem has answer yes.

479
00:29:44,120 --> 00:29:47,670
So given an NP algorithm, one
of these nondeterministic funny

480
00:29:47,670 --> 00:29:50,700
algorithms, we can convert it
into a formula satisfaction

481
00:29:50,700 --> 00:29:51,670
problem.

482
00:29:51,670 --> 00:29:53,800
And that's how you prove
3SAT is NP-complete.

483
00:29:53,800 --> 00:29:55,680
But to do that can
take many lectures,

484
00:29:55,680 --> 00:29:58,212
so I'm not going
to do the details.

485
00:30:01,195 --> 00:30:03,570
The main annoying part is
being formal about what exactly

486
00:30:03,570 --> 00:30:08,220
an algorithm is, which we
don't do in this class.

487
00:30:08,220 --> 00:30:11,820
If you're interested,
take 6.045,

488
00:30:11,820 --> 00:30:13,350
which is some
people are actually

489
00:30:13,350 --> 00:30:16,050
in the overlap this semester.

490
00:30:16,050 --> 00:30:16,550
Cool.

491
00:30:16,550 --> 00:30:17,850
Let's do some reductions.

492
00:30:17,850 --> 00:30:19,500
This is where things get fun.

493
00:30:19,500 --> 00:30:23,230
So we're going to start
with reducing 3SAT

494
00:30:23,230 --> 00:30:24,490
to Super Mario Brothers.

495
00:30:27,400 --> 00:30:31,320
So how many people have
played Super Mario Brothers?

496
00:30:31,320 --> 00:30:31,880
Easy one.

497
00:30:31,880 --> 00:30:33,769
I hope if you haven't
played, you've seen it,

498
00:30:33,769 --> 00:30:36,310
because we're going to rely very
much on Super Mario Brothers

499
00:30:36,310 --> 00:30:38,720
physics, which I hope
is fairly intuitive.

500
00:30:38,720 --> 00:30:42,050
But if you haven't played,
you should, obviously.

501
00:30:46,180 --> 00:30:50,610
And we're going to reduce
3SAT to Super Mario Brothers.

502
00:31:04,480 --> 00:31:13,440
Now this is a theorem by a
bunch of people, one MIT grad

503
00:31:13,440 --> 00:31:19,550
student, myself, and a couple
other collaborators not at MIT.

504
00:31:19,550 --> 00:31:23,220
And of course this result holds
for all versions of Super Mario

505
00:31:23,220 --> 00:31:26,569
Brothers so far
released, I think.

506
00:31:26,569 --> 00:31:28,110
The proofs are a
little bit different

507
00:31:28,110 --> 00:31:31,200
for each one, especially Mario
2, which is its own universe.

508
00:31:33,710 --> 00:31:36,430
What I'm going to talk about the
original Super Mario Brothers,

509
00:31:36,430 --> 00:31:39,380
NES classic which
I grew up with.

510
00:31:42,020 --> 00:31:45,240
Now the real Super Mario
Brothers is on a 320

511
00:31:45,240 --> 00:31:46,450
by 240 screen.

512
00:31:46,450 --> 00:31:47,480
It's a little bit small.

513
00:31:47,480 --> 00:31:51,110
Once you go right, you can't go
back left, except in the maze

514
00:31:51,110 --> 00:31:53,020
levels anyway.

515
00:31:53,020 --> 00:31:55,190
So I need to generalize
a little bit.

516
00:31:55,190 --> 00:31:58,540
Because if you assume that
the screen size of Super Mario

517
00:31:58,540 --> 00:32:00,270
Brothers is
constant, in fact you

518
00:32:00,270 --> 00:32:01,770
can dynamic program
your way through

519
00:32:01,770 --> 00:32:04,670
and find the optimal
solution in polynomial time.

520
00:32:04,670 --> 00:32:13,080
So I need to generalize a little
bit to arbitrary board size,

521
00:32:13,080 --> 00:32:15,640
arbitrary screen size.

522
00:32:15,640 --> 00:32:22,010
So in fact, my entire level will
be in one screen, no scrolling.

523
00:32:22,010 --> 00:32:24,890
Never mind this is a
side scrolling adventure.

524
00:32:24,890 --> 00:32:27,160
And so that's my
generalized problem.

525
00:32:27,160 --> 00:32:28,400
And I claim this is NP-hard.

526
00:32:28,400 --> 00:32:30,820
If I give you a
level and I ask you,

527
00:32:30,820 --> 00:32:34,070
can you get to the
end of this level?

528
00:32:34,070 --> 00:32:36,890
That problem is NP-hard.

529
00:32:36,890 --> 00:32:38,284
Also no time limit.

530
00:32:38,284 --> 00:32:40,700
The time limit would be OK,
but you have to generalize it.

531
00:32:40,700 --> 00:32:43,240
Instead of 300
seconds or whatever,

532
00:32:43,240 --> 00:32:47,119
it has to be an arbitrary value.

533
00:32:47,119 --> 00:32:48,410
So how are we going to do this?

534
00:32:48,410 --> 00:32:52,100
We're going to reduce from
3SAT to Super Mario Brothers.

535
00:32:52,100 --> 00:32:54,480
So that means I'm given--
I don't get to choose.

536
00:32:54,480 --> 00:32:56,470
I'm given one of these formulas.

537
00:32:56,470 --> 00:32:59,800
And I have to convert it into an
equivalent Super Mario Brother

538
00:32:59,800 --> 00:33:00,550
instance.

539
00:33:00,550 --> 00:33:04,300
So I have to convert it into
a level, a hypothetical level

540
00:33:04,300 --> 00:33:05,310
of Super Mario Brothers.

541
00:33:05,310 --> 00:33:08,000
Given a formula, I have
to build a level that

542
00:33:08,000 --> 00:33:11,150
implements that formula.

543
00:33:11,150 --> 00:33:13,230
So here's what it's
going to look like.

544
00:33:13,230 --> 00:33:15,300
I'm going to start
out somewhere.

545
00:33:15,300 --> 00:33:19,180
Here's my drawing of Mario.

546
00:33:19,180 --> 00:33:21,840
Mario-- or you could play Luigi.

547
00:33:21,840 --> 00:33:23,857
It doesn't matter.

548
00:33:23,857 --> 00:33:26,190
First thing it's going to do
is enter a little black box

549
00:33:26,190 --> 00:33:29,220
called a variable.

550
00:33:29,220 --> 00:33:33,800
This is supposed to
represent, let's call it x1.

551
00:33:33,800 --> 00:33:35,035
And so it's some black box.

552
00:33:35,035 --> 00:33:36,910
I'm going to tell you
what it is in a moment.

553
00:33:36,910 --> 00:33:38,180
And it has two outputs.

554
00:33:38,180 --> 00:33:41,020
There's the true output
and the false output.

555
00:33:41,020 --> 00:33:44,040
And the idea is that Mario has
to choose whether to set x1

556
00:33:44,040 --> 00:33:46,470
to true or false.

557
00:33:46,470 --> 00:33:47,676
Let me show you that gadget.

558
00:33:52,326 --> 00:33:56,880
So here's the-- whoops,
upside down-- here

559
00:33:56,880 --> 00:33:59,330
is the variable gadget.

560
00:33:59,330 --> 00:34:00,950
So here's Mario.

561
00:34:00,950 --> 00:34:02,630
Could enter from
this way or that way.

562
00:34:02,630 --> 00:34:04,610
We'll need a couple of
entrances in a moment.

563
00:34:04,610 --> 00:34:06,510
And then falls down.

564
00:34:06,510 --> 00:34:08,760
Once Mario is down here, if
you check the jump height,

565
00:34:08,760 --> 00:34:10,199
you cannot get back up to here.

566
00:34:10,199 --> 00:34:11,451
So this is like a one way.

567
00:34:11,451 --> 00:34:13,159
Once you're down here,
you have a choice.

568
00:34:13,159 --> 00:34:15,270
Should I fall to the left
or fall to the right?

569
00:34:15,270 --> 00:34:19,150
And if you make these falls
large enough, once you fall,

570
00:34:19,150 --> 00:34:21,550
you can't unfall.

571
00:34:21,550 --> 00:34:23,389
So once you make a
choice of whether I

572
00:34:23,389 --> 00:34:26,780
leave on the true exit
or the false exit,

573
00:34:26,780 --> 00:34:28,230
that's a permanent choice.

574
00:34:28,230 --> 00:34:30,730
So you can't undo it, unless
you can come back to here.

575
00:34:30,730 --> 00:34:33,487
But we'll set up so
that never happens.

576
00:34:33,487 --> 00:34:35,320
I mean, if you're trying
to solve the level,

577
00:34:35,320 --> 00:34:37,100
you don't know which way to go.

578
00:34:37,100 --> 00:34:38,212
You have to guess.

579
00:34:38,212 --> 00:34:40,170
Can I go fall to the left
or fall to the right,

580
00:34:40,170 --> 00:34:43,370
or do something.

581
00:34:43,370 --> 00:34:47,100
So the existence of a
play through, this level,

582
00:34:47,100 --> 00:34:50,870
is the same as saying there is
a choice for the x1 variable.

583
00:34:50,870 --> 00:34:53,573
Now we have to do this
for lots of variables.

584
00:34:53,573 --> 00:34:58,760
So there's x2 variable,
x3 variable, and so on.

585
00:34:58,760 --> 00:35:03,950
Each one has a true
exit and a false exit.

586
00:35:03,950 --> 00:35:06,890
So the actual level will
have n instances of this

587
00:35:06,890 --> 00:35:08,530
if we have n variables.

588
00:35:08,530 --> 00:35:11,750
Now, what do I do
once Mario decides

589
00:35:11,750 --> 00:35:14,220
that this is a true thing?

590
00:35:14,220 --> 00:35:15,680
What I'm going to
do is have-- this

591
00:35:15,680 --> 00:35:17,150
is called a gadget by the way.

592
00:35:17,150 --> 00:35:19,680
In general, most
NP-hardness proofs

593
00:35:19,680 --> 00:35:21,950
use these things called
gadgets, which is just saying,

594
00:35:21,950 --> 00:35:24,790
we take various
features of the input,

595
00:35:24,790 --> 00:35:26,940
and we convert them into
corresponding features

596
00:35:26,940 --> 00:35:27,890
on the output.

597
00:35:27,890 --> 00:35:30,770
So here I'm taking each
variable, x1, x2, x3,

598
00:35:30,770 --> 00:35:33,270
and so on, and building
this little gadget

599
00:35:33,270 --> 00:35:35,330
for each of those variables.

600
00:35:35,330 --> 00:35:39,260
Now the other main thing you
have in 3SAT are the clauses.

601
00:35:39,260 --> 00:35:42,100
We have triples of variables
or their negations.

602
00:35:42,100 --> 00:35:45,640
They have to come
together and be satisfied.

603
00:35:45,640 --> 00:35:47,320
One of them has to be true.

604
00:35:47,320 --> 00:35:56,220
So down here I'm going to have
some clause gadgets, which

605
00:35:56,220 --> 00:35:57,560
I will show you in a moment.

606
00:36:05,630 --> 00:36:08,510
OK, and I think
I'll switch colors.

607
00:36:08,510 --> 00:36:10,220
This is about to get messy.

608
00:36:10,220 --> 00:36:16,132
So the idea is that some of
the clauses have x1 in them.

609
00:36:16,132 --> 00:36:19,440
The true version
of x1, not x1 bar.

610
00:36:19,440 --> 00:36:22,920
So for those clauses,
I want to connect.

611
00:36:22,920 --> 00:36:25,010
I'm going to dip into
the clause briefly.

612
00:36:25,010 --> 00:36:27,960
So from this wire going to
dip into the clause here.

613
00:36:27,960 --> 00:36:30,950
And then I'm going to go to
the next clause that has x1.

614
00:36:30,950 --> 00:36:35,020
Maybe it's this one, and
the next one, and so on.

615
00:36:35,020 --> 00:36:38,360
All the clauses that have
x1 in it, I dip into.

616
00:36:38,360 --> 00:36:40,240
The other ones I don't.

617
00:36:40,240 --> 00:36:41,640
And then once I'm
done, I'm going

618
00:36:41,640 --> 00:36:44,290
to come back and feed into x2.

619
00:36:48,620 --> 00:36:52,180
Next, I look at this
false wire for x1.

620
00:36:52,180 --> 00:36:55,260
So all the clauses that
have x1 bar in them,

621
00:36:55,260 --> 00:36:56,720
I'm going to connect.

622
00:36:56,720 --> 00:36:58,770
So I don't know
which ones they are.

623
00:36:58,770 --> 00:37:05,420
Maybe this one, or
this one, something.

624
00:37:05,420 --> 00:37:08,270
And then I come here.

625
00:37:08,270 --> 00:37:11,150
And so the idea is that
Mario makes a choice

626
00:37:11,150 --> 00:37:12,540
whether x1 is true or false.

627
00:37:12,540 --> 00:37:17,640
If x1 is true, Mario is going
to visit all of the clauses that

628
00:37:17,640 --> 00:37:19,400
have x1 true in them.

629
00:37:19,400 --> 00:37:21,260
And then it's going to
go to the x2 choice.

630
00:37:21,260 --> 00:37:23,510
Then it's going to choose
whether x2 is true or false,

631
00:37:23,510 --> 00:37:25,620
and repeat.

632
00:37:25,620 --> 00:37:27,900
Or Mario decides
x1 should be false.

633
00:37:27,900 --> 00:37:30,420
That will satisfy
all the clauses

634
00:37:30,420 --> 00:37:34,550
that have x1 bar in them.

635
00:37:34,550 --> 00:37:37,220
And then again, we
feed back into x2.

636
00:37:37,220 --> 00:37:39,950
So this is why we have two
inputs into the x2 gadget.

637
00:37:39,950 --> 00:37:42,400
One of them is when the
previous variable was true.

638
00:37:42,400 --> 00:37:45,200
The other is when the
previous variable was false.

639
00:37:45,200 --> 00:37:47,780
The choice of x2 doesn't
depend on the choice of x1.

640
00:37:47,780 --> 00:37:49,495
So they feed into
the same thing.

641
00:37:49,495 --> 00:37:50,870
And you have to
make your choice.

642
00:37:53,960 --> 00:37:55,330
So far, so good.

643
00:37:55,330 --> 00:37:58,970
Now the question is, what's
happening in these clauses.

644
00:37:58,970 --> 00:38:03,580
And then there's
one other aspect,

645
00:38:03,580 --> 00:38:07,740
which is after you've set all of
the variables, at the very end,

646
00:38:07,740 --> 00:38:13,350
after this last variable
xn, at the very end,

647
00:38:13,350 --> 00:38:19,650
what we're going to do is come
and go through all the clauses.

648
00:38:19,650 --> 00:38:21,440
And then this is the flag.

649
00:38:21,440 --> 00:38:22,970
This is where you win the level.

650
00:38:22,970 --> 00:38:24,110
Sorry, I drew it backwards.

651
00:38:24,110 --> 00:38:29,600
But the goal is for Martin to
start here and get to here.

652
00:38:29,600 --> 00:38:31,070
In order to do
that, you have to be

653
00:38:31,070 --> 00:38:32,694
able to traverse
through these clauses.

654
00:38:32,694 --> 00:38:34,750
So what do the
clauses look like?

655
00:38:34,750 --> 00:38:37,780
This is a little
bit more elaborate.

656
00:38:37,780 --> 00:38:39,250
So here we are.

657
00:38:43,095 --> 00:38:44,970
This is a clause gadget.

658
00:38:44,970 --> 00:38:47,906
So there are three ways
to dip into the clause.

659
00:38:47,906 --> 00:38:50,030
It's actually upside down
relative to that picture,

660
00:38:50,030 --> 00:38:51,500
but that's not a problem.

661
00:38:54,180 --> 00:38:58,830
So if Mario comes here, then
he can hit the question mark

662
00:38:58,830 --> 00:38:59,850
from below.

663
00:38:59,850 --> 00:39:02,742
And inside this question mark
is an invincibility star.

664
00:39:02,742 --> 00:39:04,950
And the invincibility star
will come up here and just

665
00:39:04,950 --> 00:39:07,710
bounce around forever.

666
00:39:07,710 --> 00:39:08,610
We checked.

667
00:39:08,610 --> 00:39:12,350
The star will just stay there
for as long as you let it sit.

668
00:39:12,350 --> 00:39:14,290
Unfortunately, all of
these are solid blocks,

669
00:39:14,290 --> 00:39:18,420
so Mario can't actually get
up to here to get the star.

670
00:39:18,420 --> 00:39:20,459
But as long as Mario
can visit this question

671
00:39:20,459 --> 00:39:22,500
mark or this question mark
or this question mark,

672
00:39:22,500 --> 00:39:25,275
then there will be at
least one star up here.

673
00:39:25,275 --> 00:39:26,650
So the idea is
that each of these

674
00:39:26,650 --> 00:39:30,140
represents one of the
literals that's in the clause.

675
00:39:30,140 --> 00:39:34,560
And if we choose-- so let's look
at this first clause, x1 or x3

676
00:39:34,560 --> 00:39:35,860
or x6 bar.

677
00:39:35,860 --> 00:39:40,150
So if we choose x1 to be true,
then we'll follow the path

678
00:39:40,150 --> 00:39:42,040
and we'll be able
to hit the star.

679
00:39:42,040 --> 00:39:46,150
Or if we choose x3 to be
true, then we'll come in here

680
00:39:46,150 --> 00:39:47,910
and hit this star.

681
00:39:47,910 --> 00:39:52,199
Or if we choose x6 to
be false, then that path

682
00:39:52,199 --> 00:39:54,740
will lead to here and we'll be
able to hit this question mark

683
00:39:54,740 --> 00:39:55,800
and get the star up here.

684
00:39:55,800 --> 00:39:58,570
So as long as we
satisfy the clause,

685
00:39:58,570 --> 00:39:59,990
there will be at least one star.

686
00:39:59,990 --> 00:40:02,880
Won't help if you
have multiple stars.

687
00:40:02,880 --> 00:40:07,220
Then the final traversal part--
so that was this first clause.

688
00:40:07,220 --> 00:40:08,960
And now we're
traversing through.

689
00:40:08,960 --> 00:40:10,960
Actually in this picture,
it's left to right.

690
00:40:10,960 --> 00:40:13,420
Just turn your head.

691
00:40:13,420 --> 00:40:16,040
And so now Mario
is going to have

692
00:40:16,040 --> 00:40:19,450
to traverse this gadget from
left to right on this top part.

693
00:40:19,450 --> 00:40:22,810
And if Mario comes in here and
you can barely jump over that.

694
00:40:22,810 --> 00:40:25,120
If there's a star, you
can collect the star

695
00:40:25,120 --> 00:40:29,700
and then run through all of
these flaming bars of death.

696
00:40:29,700 --> 00:40:32,030
If there's no star, you can't.

697
00:40:32,030 --> 00:40:34,170
You'll die if you
try to traverse.

698
00:40:34,170 --> 00:40:38,530
So in order to be able to
traverse all these clauses,

699
00:40:38,530 --> 00:40:40,630
they must all be true.

700
00:40:40,630 --> 00:40:44,180
And them all being true is the
same is their AND being true.

701
00:40:44,180 --> 00:40:47,300
So you will be able to survive
through all these clauses

702
00:40:47,300 --> 00:40:51,404
if and only if this formula
has a satisfying assignment.

703
00:40:51,404 --> 00:40:52,820
The satisfying
assignment would be

704
00:40:52,820 --> 00:40:57,990
given to you by the level play.

705
00:40:57,990 --> 00:41:00,400
The choices that Mario
makes in this gadget

706
00:41:00,400 --> 00:41:02,750
will tell you
whether each variable

707
00:41:02,750 --> 00:41:05,310
should be true or false.

708
00:41:05,310 --> 00:41:08,520
So to elaborate just
a little bit more

709
00:41:08,520 --> 00:41:10,600
in general, when you have
a reduction like this,

710
00:41:10,600 --> 00:41:13,600
to prove that it actually works,
you need to check two things.

711
00:41:13,600 --> 00:41:15,690
You need to check
that if there is a way

712
00:41:15,690 --> 00:41:17,670
to satisfy this
formula, then there

713
00:41:17,670 --> 00:41:19,300
is a way to play this level.

714
00:41:19,300 --> 00:41:21,716
And then conversely you need
to show that if there's a way

715
00:41:21,716 --> 00:41:23,610
to play this level,
then the formula

716
00:41:23,610 --> 00:41:25,150
has a satisfying assignment.

717
00:41:25,150 --> 00:41:28,980
So for that latter
part, in order

718
00:41:28,980 --> 00:41:31,437
to convert a level play into
a satisfying assignment,

719
00:41:31,437 --> 00:41:34,020
you just check which way Mario
falls in each of these gadgets,

720
00:41:34,020 --> 00:41:34,820
left or right.

721
00:41:34,820 --> 00:41:36,700
That tells you the
variable assignment.

722
00:41:36,700 --> 00:41:38,860
And because of the
way the clauses work,

723
00:41:38,860 --> 00:41:41,200
you'll only be able to
finish the level if there

724
00:41:41,200 --> 00:41:42,910
was at least one star here.

725
00:41:42,910 --> 00:41:44,780
And stars run out
after some time.

726
00:41:44,780 --> 00:41:48,530
So you can barely make it
through all the flaming bars

727
00:41:48,530 --> 00:41:49,067
of death.

728
00:41:49,067 --> 00:41:50,400
Then you get to the next clause.

729
00:41:50,400 --> 00:41:53,230
You need another
star for each one.

730
00:41:53,230 --> 00:41:56,300
Conversely, if there is
a satisfying assignment,

731
00:41:56,300 --> 00:41:58,060
you can actually play
through the level,

732
00:41:58,060 --> 00:41:59,934
you just make these
choices according to what

733
00:41:59,934 --> 00:42:01,180
the satisfying assignment is.

734
00:42:01,180 --> 00:42:03,330
So either way it's equivalent.

735
00:42:03,330 --> 00:42:07,090
We always get a yes or
no answer here whenever

736
00:42:07,090 --> 00:42:11,460
we get a corresponding yes or
no answer to the 3SAT process.

737
00:42:11,460 --> 00:42:15,330
You also need to check that this
reduction is polynomial size.

738
00:42:15,330 --> 00:42:17,690
It can be computed
in polynomial time.

739
00:42:17,690 --> 00:42:18,530
So there's an issue.

740
00:42:18,530 --> 00:42:21,420
Given this thing, you have
to lay this out in a grid

741
00:42:21,420 --> 00:42:23,780
and draw all these wires.

742
00:42:23,780 --> 00:42:28,290
And there's one
problem here, which is,

743
00:42:28,290 --> 00:42:31,010
these wires cross each other.

744
00:42:31,010 --> 00:42:33,870
And that's a little
awkward, because these wires

745
00:42:33,870 --> 00:42:36,660
are basically just long tunnels
for Mario to walk through.

746
00:42:36,660 --> 00:42:38,910
But what does it mean
to have a crossing wire?

747
00:42:38,910 --> 00:42:41,400
Really, if Mario's
coming this way,

748
00:42:41,400 --> 00:42:44,630
I don't want them to
be able to go up here.

749
00:42:44,630 --> 00:42:46,100
He has to go straight.

750
00:42:46,100 --> 00:42:47,900
Otherwise this
reduction won't work.

751
00:42:47,900 --> 00:42:50,800
So I need what's called
a crossover gadget.

752
00:42:50,800 --> 00:42:56,940
And everywhere here I have a
crossing, I have a crossover.

753
00:42:56,940 --> 00:42:59,360
And this gadget has
to guarantee that I

754
00:42:59,360 --> 00:43:01,220
can go through one
way or the other way,

755
00:43:01,220 --> 00:43:04,860
but there's no leakage from
one path to the other path.

756
00:43:04,860 --> 00:43:08,550
Actually, if I first
traverse through here,

757
00:43:08,550 --> 00:43:10,950
and then I traverse through
here, it's OK if I leak back.

758
00:43:10,950 --> 00:43:14,310
Because once I visit a
wire, it's kind of done.

759
00:43:14,310 --> 00:43:17,430
But I can't have leakage if
only one of them is traversed.

760
00:43:17,430 --> 00:43:23,570
So this is the last gadget, the
most complicated of them all.

761
00:43:23,570 --> 00:43:25,295
So this took a
while to construct,

762
00:43:25,295 --> 00:43:26,170
as you might imagine.

763
00:43:28,710 --> 00:43:32,470
So this is what we call a
unidirectional crossover.

764
00:43:32,470 --> 00:43:38,690
You can either go from left to
right or from bottom to top,

765
00:43:38,690 --> 00:43:42,270
but you cannot go from bottom to
right or bottom to left or left

766
00:43:42,270 --> 00:43:44,480
to bottom, that kind of thing.

767
00:43:44,480 --> 00:43:45,980
So I'm told that
Mario is only going

768
00:43:45,980 --> 00:43:48,450
to enter from here to here,
because all of these wires,

769
00:43:48,450 --> 00:43:50,010
I can make one way wires.

770
00:43:50,010 --> 00:43:53,440
I only have to think about
going in a particular direction.

771
00:43:53,440 --> 00:43:55,620
I can have falls to
force Mario to only go

772
00:43:55,620 --> 00:43:57,840
one way along these wires.

773
00:43:57,840 --> 00:44:01,790
And so let me show you
the valid traversals.

774
00:44:01,790 --> 00:44:03,900
Maybe the simplest
one is from here.

775
00:44:03,900 --> 00:44:05,910
So let's say Mario
comes in here, falls.

776
00:44:05,910 --> 00:44:08,480
So I can't backtrack,
can jump up here.

777
00:44:08,480 --> 00:44:11,960
And then if Mario's big,
he can break this block,

778
00:44:11,960 --> 00:44:12,890
break this block.

779
00:44:12,890 --> 00:44:17,490
But if he's big-- there should
be a couple more zig zags here.

780
00:44:17,490 --> 00:44:18,790
Let's try to run.

781
00:44:18,790 --> 00:44:21,729
You can crouch
slide through here.

782
00:44:21,729 --> 00:44:23,520
But then you'll sort
of lose your momentum,

783
00:44:23,520 --> 00:44:25,144
and you won't be able
to go through all

784
00:44:25,144 --> 00:44:27,990
these traversals as big Mario.

785
00:44:27,990 --> 00:44:31,800
So you can break these blocks
and then get up to the top

786
00:44:31,800 --> 00:44:32,800
and leave.

787
00:44:32,800 --> 00:44:35,520
Or, if big Mario comes
from over this way,

788
00:44:35,520 --> 00:44:38,940
you can first take a
damage, become small Mario.

789
00:44:38,940 --> 00:44:42,010
Then you can fit through
these wiggly blocks.

790
00:44:42,010 --> 00:44:45,090
But you cannot break blocks
anymore as small Mario.

791
00:44:45,090 --> 00:44:48,230
So once you've committed
to going small,

792
00:44:48,230 --> 00:44:50,520
you have to stay small,
until you get to here.

793
00:44:50,520 --> 00:44:52,402
And then there's a
mushroom in this block.

794
00:44:52,402 --> 00:44:54,860
So you can get big again, and
then you can break this block

795
00:44:54,860 --> 00:44:56,020
and leave.

796
00:44:56,020 --> 00:44:57,700
But once you're big,
you can't backtrack

797
00:44:57,700 --> 00:45:00,690
because big Mario can't fit
through these tiny tubes.

798
00:45:00,690 --> 00:45:03,430
See it clear, right?

799
00:45:03,430 --> 00:45:07,260
So slight detail, which
is at the beginning,

800
00:45:07,260 --> 00:45:10,660
we need to make Mario big--
so there's a little mushroom.

801
00:45:10,660 --> 00:45:14,450
I think they have three
spots-- at the beginning.

802
00:45:14,450 --> 00:45:16,180
And also at the
end, there has to be

803
00:45:16,180 --> 00:45:18,600
something like this that
checks that you actually

804
00:45:18,600 --> 00:45:19,640
have a mushroom.

805
00:45:19,640 --> 00:45:22,180
So the only time you're
allowed to take damage

806
00:45:22,180 --> 00:45:24,232
is briefly in this
gadget you take damage.

807
00:45:24,232 --> 00:45:26,190
If you tried to backtrack,
you would get stuck.

808
00:45:26,190 --> 00:45:28,120
There's a long fall here.

809
00:45:28,120 --> 00:45:29,820
And then you have
to get the mushroom

810
00:45:29,820 --> 00:45:32,610
so you can escape again.

811
00:45:32,610 --> 00:45:39,250
So at the end there's
like a mushroom check.

812
00:45:39,250 --> 00:45:40,950
Make sure you have it.

813
00:45:40,950 --> 00:45:42,790
So most of the
time Mario is big.

814
00:45:42,790 --> 00:45:44,260
And just in these
little crossovers

815
00:45:44,260 --> 00:45:45,635
you have to make
these decisions.

816
00:45:45,635 --> 00:45:47,740
This would make a
giant level, but it

817
00:45:47,740 --> 00:45:54,650
is polynomial size, probably
quadratic or something.

818
00:45:54,650 --> 00:45:56,400
Therefore Super Mario
Brothers is NP-hard.

819
00:45:59,050 --> 00:46:01,420
So if you want more
fun examples like this,

820
00:46:01,420 --> 00:46:04,650
you should check out 6.890, the
class I taught last semester,

821
00:46:04,650 --> 00:46:08,100
which has online video lectures,
soon to be on OpenCourseWare.

822
00:46:08,100 --> 00:46:12,310
So you can play with that.

823
00:46:12,310 --> 00:46:14,310
Any questions about Mario?

824
00:46:17,340 --> 00:46:18,740
All right, I hope you all play.

825
00:46:27,380 --> 00:46:30,730
So the next topic is
a problem you probably

826
00:46:30,730 --> 00:46:33,410
haven't heard about, three
dimensional matching.

827
00:46:43,880 --> 00:46:46,980
This is a kind of a
graph theory problem.

828
00:46:46,980 --> 00:46:50,590
We're going to call
it 3DM for short.

829
00:46:50,590 --> 00:46:53,640
And you've seen matching
problems based on flow.

830
00:46:53,640 --> 00:46:56,030
Matching problems are usually
about pairs of things.

831
00:46:56,030 --> 00:46:58,220
You're pairing them
up, which you might

832
00:46:58,220 --> 00:46:59,470
call two dimensional matching.

833
00:46:59,470 --> 00:47:01,700
That can be solved
in polynomial time.

834
00:47:01,700 --> 00:47:05,380
But if you change two to three
and you're tripling things up,

835
00:47:05,380 --> 00:47:08,420
then suddenly the problem
becomes NP-complete.

836
00:47:08,420 --> 00:47:13,290
So it's a useful starting
point, similar to 3SAT.

837
00:47:20,090 --> 00:47:28,250
So you're given a
set X of elements,

838
00:47:28,250 --> 00:47:30,810
a set Y of elements,
a set Z of elements.

839
00:47:30,810 --> 00:47:32,770
None of them are shared.

840
00:47:32,770 --> 00:47:37,020
But more importantly, you
are given a bunch of triples.

841
00:47:37,020 --> 00:47:40,850
These are the allowable triples.

842
00:47:40,850 --> 00:47:44,790
So we'll call the set
of allowable triples T.

843
00:47:44,790 --> 00:47:47,100
And so we're looking
at the cross product.

844
00:47:47,100 --> 00:47:50,890
This is the set of all triples
X, Y, and Z, or X is in X,

845
00:47:50,890 --> 00:47:54,620
Y is in Y, and Z is in Z. But
not all triples are allowed.

846
00:47:54,620 --> 00:47:58,210
Only some subset of
triples is allowed.

847
00:47:58,210 --> 00:48:04,800
And your goal is to choose
among those subsets-- sorry,

848
00:48:04,800 --> 00:48:09,440
among those triples a
subset of the triples.

849
00:48:09,440 --> 00:48:11,020
So we're trying
to choose a subset

850
00:48:11,020 --> 00:48:26,667
S of T such that every element--
so the things in X, Y, and Z

851
00:48:26,667 --> 00:48:27,500
are called elements.

852
00:48:27,500 --> 00:48:33,240
So I'm just taking
somebody in the union XYZ.

853
00:48:33,240 --> 00:48:48,225
It should be in exactly
one triple s in big S.

854
00:48:48,225 --> 00:48:50,600
This is a little weird, but
you can think of this problem

855
00:48:50,600 --> 00:48:56,140
as you have an alien
race with three genders--

856
00:48:56,140 --> 00:48:58,254
male, female, neuter I guess.

857
00:48:58,254 --> 00:48:59,420
Those are the X, Y, and Z's.

858
00:48:59,420 --> 00:49:02,420
There's an equal number of each.

859
00:49:02,420 --> 00:49:06,010
And every triple reports
to you whether that

860
00:49:06,010 --> 00:49:08,270
is a compatible matching.

861
00:49:08,270 --> 00:49:11,130
Who knows what they're
doing, all three of them?

862
00:49:11,130 --> 00:49:14,110
So you're told up front--
you take a survey.

863
00:49:14,110 --> 00:49:15,790
There's only n cubed
different triples.

864
00:49:15,790 --> 00:49:19,992
For each of them they
say, yeah, I'd do that.

865
00:49:19,992 --> 00:49:22,630
So you were given that subset.

866
00:49:22,630 --> 00:49:26,400
And now your goal is to
permanently triple up

867
00:49:26,400 --> 00:49:27,380
these guys.

868
00:49:27,380 --> 00:49:30,630
And everybody wants to
be in exactly one triple.

869
00:49:30,630 --> 00:49:34,860
So it's a monogamous
race, imagine.

870
00:49:34,860 --> 00:49:38,190
So everybody wants to be put
in one triple, but only one

871
00:49:38,190 --> 00:49:39,317
triple.

872
00:49:39,317 --> 00:49:40,900
And the question is,
is this possible?

873
00:49:40,900 --> 00:49:42,358
This is three
dimensional matching.

874
00:49:42,358 --> 00:49:45,436
Certainly not always going to be
possible, but sometimes it is.

875
00:49:45,436 --> 00:49:46,810
If it is, you want
to answer yes.

876
00:49:46,810 --> 00:49:50,060
If it's not possible,
you want to answer no.

877
00:49:50,060 --> 00:49:52,180
This problem is NP-complete.

878
00:49:52,180 --> 00:49:54,150
Why is it in NP?

879
00:49:54,150 --> 00:49:58,140
Because I can basically guess
which elements of T are in S.

880
00:49:58,140 --> 00:50:00,730
There's only at most
n cubed of them.

881
00:50:00,730 --> 00:50:02,510
So for each one, it
is guess yes or no,

882
00:50:02,510 --> 00:50:04,310
is that element of T in S?

883
00:50:04,310 --> 00:50:07,270
And then I check whether this
coverage constraint holds.

884
00:50:07,270 --> 00:50:09,617
So it's very easy to
prove this is in NP.

885
00:50:12,540 --> 00:50:15,940
The challenge is to
prove that it's NP-hard.

886
00:50:15,940 --> 00:50:20,610
And we're going to do that,
again, by reducing from 3SAT.

887
00:50:32,510 --> 00:50:36,640
So we're going to make
a reduction from 3SAT

888
00:50:36,640 --> 00:50:38,850
to three dimensional matching.

889
00:50:38,850 --> 00:50:39,840
Direction is important.

890
00:50:39,840 --> 00:50:42,030
Always reduce from the
thing you know is hard

891
00:50:42,030 --> 00:50:45,230
and reduce to the thing
you don't know is hard.

892
00:50:45,230 --> 00:50:47,320
So again, we're given a formula.

893
00:50:47,320 --> 00:50:49,640
And we want to
convert that formula

894
00:50:49,640 --> 00:50:53,670
into an equivalent three
dimensional matching input.

895
00:50:53,670 --> 00:50:56,965
So the formula has
variables and clauses.

896
00:50:56,965 --> 00:50:58,590
For each variable,
we're going to build

897
00:50:58,590 --> 00:51:01,777
a gadget that looks like this.

898
00:51:01,777 --> 00:51:03,860
And for each clause we're
going to build a gadget.

899
00:51:03,860 --> 00:51:06,720
So here's what they look like.

900
00:51:06,720 --> 00:51:13,730
If we have a variable x1,
we're going to convert that

901
00:51:13,730 --> 00:51:15,650
into this picture.

902
00:51:30,490 --> 00:51:32,053
Stay monochromatic for now.

903
00:51:36,660 --> 00:51:41,176
Looks pretty crazy at the
moment, but it's not so crazy.

904
00:51:56,510 --> 00:51:58,490
This is not supposed
to be obvious.

905
00:51:58,490 --> 00:52:00,330
You have to think for a while.

906
00:52:00,330 --> 00:52:03,310
It's a puzzle to figure
out this kind of thing.

907
00:52:03,310 --> 00:52:08,610
But I call this thing a variable
gadget because locally--

908
00:52:08,610 --> 00:52:11,450
so there's basically a
wheel in the center here.

909
00:52:11,450 --> 00:52:13,740
And then there's
these extra dots

910
00:52:13,740 --> 00:52:15,810
for every pair of dots,
consecutive pairs of dots

911
00:52:15,810 --> 00:52:17,100
in a wheel.

912
00:52:17,100 --> 00:52:19,960
And what I've drawn is the set
of triples that are allowed.

913
00:52:19,960 --> 00:52:22,620
There's tons of other
triples which are forbidden.

914
00:52:22,620 --> 00:52:24,670
The triples that are
in T are the ones

915
00:52:24,670 --> 00:52:27,490
that I draw as little triangles.

916
00:52:27,490 --> 00:52:30,480
And two color them because
there are exactly two ways

917
00:52:30,480 --> 00:52:31,850
to solve this gadget locally.

918
00:52:31,850 --> 00:52:35,310
Now these dots are going to
be connected to other gadgets.

919
00:52:35,310 --> 00:52:39,407
But these dots only exist
in this gadget, which means

920
00:52:39,407 --> 00:52:40,490
they've got to be covered.

921
00:52:40,490 --> 00:52:42,115
They've got to be
covered exactly once.

922
00:52:42,115 --> 00:52:45,470
So either you choose
the blue triangles,

923
00:52:45,470 --> 00:52:47,050
or you choose the red triangles.

924
00:52:47,050 --> 00:52:51,840
Each of them will exactly
cover each of these guys once.

925
00:52:51,840 --> 00:52:53,890
You cannot mix and
match red and blue,

926
00:52:53,890 --> 00:52:56,810
because you either get overlap
if you choose two guys that

927
00:52:56,810 --> 00:52:59,250
share a point, or
you'd miss one.

928
00:52:59,250 --> 00:53:01,770
If I choose like this
blue and this red,

929
00:53:01,770 --> 00:53:04,550
then I can't cover this
point because both of these

930
00:53:04,550 --> 00:53:05,910
would overlap those two.

931
00:53:05,910 --> 00:53:09,080
And over here you have to
choose [INAUDIBLE] triples.

932
00:53:09,080 --> 00:53:10,590
They can't overlap at all.

933
00:53:10,590 --> 00:53:13,270
And everybody has
to get covered.

934
00:53:13,270 --> 00:53:15,640
So just given those
constraints, locally you

935
00:53:15,640 --> 00:53:18,260
can see you have to
choose red or blue.

936
00:53:18,260 --> 00:53:18,920
Guess what?

937
00:53:18,920 --> 00:53:21,800
One of them is true,
the other one is false.

938
00:53:21,800 --> 00:53:25,414
Let's say that red is
true and blue is false.

939
00:53:25,414 --> 00:53:27,830
In general, when you're trying
to build a variable gadget,

940
00:53:27,830 --> 00:53:30,180
you build something
that has exactly

941
00:53:30,180 --> 00:53:33,220
two solutions, one representing
true, one representing false.

942
00:53:33,220 --> 00:53:36,850
Now how big do I
make this wheel?

943
00:53:36,850 --> 00:53:39,800
Big enough.

944
00:53:39,800 --> 00:53:41,970
You could make it as big
as the number of clauses.

945
00:53:41,970 --> 00:53:46,890
I'm going to make
it into two and x1.

946
00:53:46,890 --> 00:53:59,110
So wheel-- and this number
is the number of occurrences

947
00:53:59,110 --> 00:54:03,680
of x1 in the formula.

948
00:54:03,680 --> 00:54:06,900
So this is the number of clauses
that contain either xi or xi

949
00:54:06,900 --> 00:54:08,720
bar.

950
00:54:08,720 --> 00:54:09,585
That's in xi.

951
00:54:09,585 --> 00:54:11,580
I'm going to double that.

952
00:54:11,580 --> 00:54:14,960
Because what I get over
here is basically xi

953
00:54:14,960 --> 00:54:21,540
being true for those guys.

954
00:54:21,540 --> 00:54:25,020
Actually, yeah,
that's actually right.

955
00:54:25,020 --> 00:54:27,220
It looks backwards.

956
00:54:27,220 --> 00:54:28,650
And false for these guys.

957
00:54:31,670 --> 00:54:35,380
One way or the other,
we'll figure it out.

958
00:54:35,380 --> 00:54:40,800
So in order for xi to appear
in, say, five different clauses,

959
00:54:40,800 --> 00:54:45,590
I want five of the true things
and five of the false things.

960
00:54:45,590 --> 00:54:49,854
And so I need to double in
order to get-- potentially

961
00:54:49,854 --> 00:54:51,520
I have twice as many
as I actually need,

962
00:54:51,520 --> 00:54:53,603
but this way I'm guaranteed
to have false or true,

963
00:54:53,603 --> 00:54:55,830
whichever I need.

964
00:54:55,830 --> 00:54:57,560
In reality I have
some true occurrences.

965
00:54:57,560 --> 00:55:01,210
I have some false occurrences,
some x1's, some x1 bars.

966
00:55:01,210 --> 00:55:06,600
This will guarantee that I have
enough of these free points

967
00:55:06,600 --> 00:55:09,720
to connect into
my clause gadgets.

968
00:55:09,720 --> 00:55:11,359
How do I do a clause gadget?

969
00:55:11,359 --> 00:55:12,442
It's actually really easy.

970
00:55:16,854 --> 00:55:18,770
So these would be pretty
boring by themselves.

971
00:55:25,590 --> 00:55:27,180
So a clause always
looks like this.

972
00:55:27,180 --> 00:55:28,388
Maybe there's some negations.

973
00:55:30,860 --> 00:55:34,620
Yeah, let's do
something like that.

974
00:55:34,620 --> 00:55:36,795
I'm going to convert it
into a very simple picture.

975
00:55:43,810 --> 00:55:50,700
It's going to be xi dot,
and xj bar dot, and xk dot.

976
00:55:50,700 --> 00:55:56,363
And then-- well maybe I'll
stick to these colors.

977
00:56:02,400 --> 00:56:06,940
Again, these two points only
appear in this clause gadget.

978
00:56:06,940 --> 00:56:10,410
These dots are
actually these dots.

979
00:56:10,410 --> 00:56:12,570
So there's one of
these pictures for x1.

980
00:56:12,570 --> 00:56:14,330
There's another one for x2, x3.

981
00:56:14,330 --> 00:56:16,770
And so xi has one
of these wheels.

982
00:56:16,770 --> 00:56:20,190
I want this dot to be one
of these dots of the wheel.

983
00:56:20,190 --> 00:56:22,900
And then I want
this dot to be one

984
00:56:22,900 --> 00:56:26,300
of the dots in the xj wheel
with the false setting, one

985
00:56:26,300 --> 00:56:27,600
of the red dots.

986
00:56:27,600 --> 00:56:34,981
I want this one to be xk
true setting in the xk wheel.

987
00:56:34,981 --> 00:56:36,730
So these things are
all connected together

988
00:56:36,730 --> 00:56:38,310
in a complicated pattern.

989
00:56:38,310 --> 00:56:40,110
But the point is that
within this gadget,

990
00:56:40,110 --> 00:56:42,910
I only have three
allowed triples.

991
00:56:42,910 --> 00:56:44,790
And these points only
appear in this gadget,

992
00:56:44,790 --> 00:56:47,734
which means they have to
be covered in this gadget.

993
00:56:47,734 --> 00:56:49,150
They can be covered
by this triple

994
00:56:49,150 --> 00:56:51,200
or this triple or this triple.

995
00:56:51,200 --> 00:56:54,480
But once you choose one,
you can't choose the others.

996
00:56:54,480 --> 00:56:59,540
What this means is if
I set x1 to be true,

997
00:56:59,540 --> 00:57:03,540
it leaves behind these
points marked true.

998
00:57:03,540 --> 00:57:05,640
If I choose the red
things, then it's

999
00:57:05,640 --> 00:57:08,184
the blue points that
are left behind.

1000
00:57:08,184 --> 00:57:09,600
Leaving points
behind in this case

1001
00:57:09,600 --> 00:57:11,772
is going to be good,
because this clause,

1002
00:57:11,772 --> 00:57:13,480
in order to satisfy
this clause, in order

1003
00:57:13,480 --> 00:57:17,330
to choose one of these three
triples, at least one of these

1004
00:57:17,330 --> 00:57:22,500
must be left behind
by the wheel.

1005
00:57:22,500 --> 00:57:24,457
If all of these are
covered by their wheels,

1006
00:57:24,457 --> 00:57:25,290
then there's no way.

1007
00:57:25,290 --> 00:57:27,750
I can't choose
any of these guys.

1008
00:57:27,750 --> 00:57:30,810
But if at least one of these
is left behind by the wheel,

1009
00:57:30,810 --> 00:57:33,780
then I can choose the
corresponding triple

1010
00:57:33,780 --> 00:57:34,760
and cover these points.

1011
00:57:34,760 --> 00:57:36,880
So I'll be able to cover
these points if and only

1012
00:57:36,880 --> 00:57:39,947
if at least one
of these is true.

1013
00:57:39,947 --> 00:57:40,780
And that's a clause.

1014
00:57:40,780 --> 00:57:43,440
That's what a clause is
supposed to do in 3SAT.

1015
00:57:43,440 --> 00:57:45,200
If at least one
of these is true,

1016
00:57:45,200 --> 00:57:46,590
then the clause is satisfied.

1017
00:57:46,590 --> 00:57:48,380
I need all the clauses
to be satisfied

1018
00:57:48,380 --> 00:57:50,720
because I need to cover
of these points for all

1019
00:57:50,720 --> 00:57:53,760
the instances of these clauses.

1020
00:57:53,760 --> 00:57:55,310
And that's how it works.

1021
00:57:55,310 --> 00:57:57,540
Now, slight catch.

1022
00:57:57,540 --> 00:57:59,710
If you do this,
not all the points

1023
00:57:59,710 --> 00:58:01,740
will be covered, even so.

1024
00:58:04,650 --> 00:58:05,970
Maybe all of these are true.

1025
00:58:05,970 --> 00:58:07,890
And so they're all left behind.

1026
00:58:07,890 --> 00:58:10,580
And I can only cover one
of them with the clause.

1027
00:58:10,580 --> 00:58:12,560
It's a little messy.

1028
00:58:12,560 --> 00:58:17,710
You need another gadget, which
is called garbage collection.

1029
00:58:23,910 --> 00:58:26,650
I don't want to spend
too much time on it.

1030
00:58:26,650 --> 00:58:30,900
But you have two dots.

1031
00:58:30,900 --> 00:58:42,610
And then you have every
single xi-- these dots,

1032
00:58:42,610 --> 00:58:47,490
all true and false dots.

1033
00:58:47,490 --> 00:58:49,410
And you're going to
have this triple,

1034
00:58:49,410 --> 00:58:52,690
and this triple, and this
triple, and this triple,

1035
00:58:52,690 --> 00:58:53,220
and so on.

1036
00:58:53,220 --> 00:58:56,230
It looks an awful
lot like a clause.

1037
00:58:56,230 --> 00:58:59,190
But this is like a clause
that's connected to everybody

1038
00:58:59,190 --> 00:59:00,820
in the entire universe.

1039
00:59:00,820 --> 00:59:03,260
And you repeat this
the appropriate number

1040
00:59:03,260 --> 00:59:09,380
of times, which is
something like sum of nx

1041
00:59:09,380 --> 00:59:10,670
minus the number of clauses.

1042
00:59:14,100 --> 00:59:14,850
OK, why?

1043
00:59:14,850 --> 00:59:16,880
Because if you
look at a wheel, it

1044
00:59:16,880 --> 00:59:21,780
has size 2 times nx
for a variable x.

1045
00:59:21,780 --> 00:59:26,300
And half of the points
will be left uncovered.

1046
00:59:26,300 --> 00:59:29,630
So that means nx of
them will be uncovered.

1047
00:59:29,630 --> 00:59:33,110
Then the clause, if everything
works out correctly,

1048
00:59:33,110 --> 00:59:37,710
the clause will cover
exactly one of those points.

1049
00:59:37,710 --> 00:59:40,090
So for each clause we
cover one of the points.

1050
00:59:40,090 --> 00:59:43,270
That means this
difference is exactly how

1051
00:59:43,270 --> 00:59:45,470
many points are left uncovered.

1052
00:59:45,470 --> 00:59:48,720
And so we make this gadget
exactly that many times.

1053
00:59:48,720 --> 00:59:50,530
And it's free to cover anybody.

1054
00:59:50,530 --> 00:59:53,860
So whatever is left over,
this garbage collector

1055
00:59:53,860 --> 00:59:55,160
will clean up.

1056
00:59:55,160 --> 00:59:57,200
And if we use exactly
the right number of them,

1057
00:59:57,200 --> 01:00:01,170
this garbage collector won't
run out of things to collect.

1058
01:00:01,170 --> 01:00:03,830
So this makes the proof messy.

1059
01:00:03,830 --> 01:00:07,712
But I want to move on to
somewhat simpler proofs

1060
01:00:07,712 --> 01:00:08,670
and for other problems.

1061
01:00:08,670 --> 01:00:09,832
Yeah?

1062
01:00:09,832 --> 01:00:12,262
AUDIENCE: Real quick,
what about the t or f

1063
01:00:12,262 --> 01:00:14,692
points that we didn't cover
because we didn't actually

1064
01:00:14,692 --> 01:00:15,670
need that many?

1065
01:00:15,670 --> 01:00:16,503
ERIC DEMAINE: Right.

1066
01:00:16,503 --> 01:00:19,210
So this also includes the points
that weren't even connected

1067
01:00:19,210 --> 01:00:20,950
to clauses.

1068
01:00:20,950 --> 01:00:23,330
I think this is the right
number no matter what,

1069
01:00:23,330 --> 01:00:26,180
because this is counting the
total number of uncovered guys,

1070
01:00:26,180 --> 01:00:28,310
whether they're connected
to clauses or not.

1071
01:00:28,310 --> 01:00:30,890
Each clause will, in
a satisfied situation,

1072
01:00:30,890 --> 01:00:33,341
it will cover exactly
one of those points.

1073
01:00:33,341 --> 01:00:35,090
The ones that are
connected to the clauses

1074
01:00:35,090 --> 01:00:37,010
won't be covered at
all, but that will still

1075
01:00:37,010 --> 01:00:38,000
be in this difference.

1076
01:00:38,000 --> 01:00:39,750
So yeah, it's good
to check that.

1077
01:00:39,750 --> 01:00:42,250
The first time I wrote this
down I forgot about those points

1078
01:00:42,250 --> 01:00:42,958
and got it wrong.

1079
01:00:42,958 --> 01:00:45,930
But I think this is
right, hopefully.

1080
01:00:45,930 --> 01:00:47,980
I did not come up
with this proof.

1081
01:00:47,980 --> 01:00:49,950
Garey and Johnson
I think-- or no.

1082
01:00:49,950 --> 01:00:52,906
This is-- I forgot.

1083
01:00:52,906 --> 01:00:54,710
Yeah, this is a Garey
and Johnson proof.

1084
01:00:54,710 --> 01:00:58,610
There's a cool book from the
late '70s by Garey and Johnson,

1085
01:00:58,610 --> 01:01:04,320
does a lot of NP-completeness,
if you're curious.

1086
01:01:04,320 --> 01:01:06,702
All right, so
hopefully you believe

1087
01:01:06,702 --> 01:01:08,160
three dimensional
matching is hard.

1088
01:01:08,160 --> 01:01:10,520
Now I'm going to use it to prove
that some very different types

1089
01:01:10,520 --> 01:01:11,650
of problems are hard.

1090
01:01:11,650 --> 01:01:14,270
This is a kind of
graph theory problem.

1091
01:01:14,270 --> 01:01:18,070
You'll see more graph theory
problems in recitation.

1092
01:01:18,070 --> 01:01:21,800
This one, I can
erase 3SAT and Mario.

1093
01:01:21,800 --> 01:01:27,660
So in the world, most
NP-hardness proofs

1094
01:01:27,660 --> 01:01:31,520
are reductions from 3SAT,
or some variation of 3SAT.

1095
01:01:31,520 --> 01:01:34,450
In some sense, you can think
of three dimensional matching

1096
01:01:34,450 --> 01:01:37,870
as kind of like a
version of 3SAT,

1097
01:01:37,870 --> 01:01:39,840
but it's a little
bit more stringent.

1098
01:01:39,840 --> 01:01:44,980
And that stringency helps
us to do other reductions.

1099
01:01:44,980 --> 01:01:47,070
So here's another
problem where we'll

1100
01:01:47,070 --> 01:01:51,710
reduce from three
dimensional matching.

1101
01:01:51,710 --> 01:01:52,940
It's called subset sum.

1102
01:02:00,220 --> 01:02:09,830
So you're given n
integers, a1 up to an.

1103
01:02:09,830 --> 01:02:15,090
And you're given a target
sum, also an integer.

1104
01:02:15,090 --> 01:02:17,150
Call it t.

1105
01:02:17,150 --> 01:02:21,160
What you'd like to know is, is
there a subset of the integers

1106
01:02:21,160 --> 01:02:22,550
that adds up to that target.

1107
01:02:34,380 --> 01:02:36,230
Can you choose a
sum of the integers

1108
01:02:36,230 --> 01:02:39,800
so that-- I'll write
it the sum of S.

1109
01:02:39,800 --> 01:02:43,670
But what this means is
the sum over the ai's that

1110
01:02:43,670 --> 01:02:47,576
are in S of the value ai.

1111
01:02:47,576 --> 01:02:51,540
I want that to equal t.

1112
01:02:51,540 --> 01:02:52,980
So this is the definition.

1113
01:02:52,980 --> 01:02:55,206
This is the constraint.

1114
01:02:55,206 --> 01:02:56,580
So I give you a
bunch of numbers.

1115
01:02:56,580 --> 01:02:58,480
Do any subset of
them add up to t?

1116
01:02:58,480 --> 01:02:59,720
That's all this is asking.

1117
01:02:59,720 --> 01:03:02,740
This problem is NP-hard.

1118
01:03:02,740 --> 01:03:06,265
It's NP-complete, in fact, when
you can guess which integers

1119
01:03:06,265 --> 01:03:08,140
should go in the subset,
and then add them up

1120
01:03:08,140 --> 01:03:09,264
to see if you got it right.

1121
01:03:11,886 --> 01:03:14,320
It is NP-hard, but
it's something special

1122
01:03:14,320 --> 01:03:15,960
we call weakly NP-hard.

1123
01:03:22,430 --> 01:03:28,687
And why don't I come back to the
definition of that in a moment?

1124
01:03:28,687 --> 01:03:30,020
Let me first show you the proof.

1125
01:03:30,020 --> 01:03:33,510
It's actually really easy
now that we have this three

1126
01:03:33,510 --> 01:03:37,120
dimensional matching problem.

1127
01:03:37,120 --> 01:03:38,820
It's pretty cool.

1128
01:03:38,820 --> 01:03:42,530
So these numbers are
going to be huge.

1129
01:03:42,530 --> 01:03:48,600
What we're going to say
is, let's view-- so again,

1130
01:03:48,600 --> 01:03:51,587
we're given a three
dimensional matching instance.

1131
01:03:51,587 --> 01:03:52,670
Get the directions, right?

1132
01:03:52,670 --> 01:03:54,520
We're given a set of triples.

1133
01:03:54,520 --> 01:03:59,520
We want to solve this problem
by reducing it to a subset sum.

1134
01:03:59,520 --> 01:04:04,688
So we get to construct integers
that represent triples.

1135
01:04:04,688 --> 01:04:06,220
That's what we're going to do.

1136
01:04:06,220 --> 01:04:09,477
So here we go.

1137
01:04:09,477 --> 01:04:10,560
We get to choose a number.

1138
01:04:10,560 --> 01:04:16,440
So I'm going to think of
them in a particular base, b,

1139
01:04:16,440 --> 01:04:23,420
which is going to be 1
plus the max of the mxi's.

1140
01:04:23,420 --> 01:04:26,310
So again, this is the number
of occurrences of variable xi

1141
01:04:26,310 --> 01:04:28,610
in a true or false form.

1142
01:04:28,610 --> 01:04:31,490
So I take the maximum occurrence
of any variable, add 1.

1143
01:04:31,490 --> 01:04:32,490
That's my base.

1144
01:04:32,490 --> 01:04:34,170
It just has to be large enough.

1145
01:04:37,010 --> 01:04:43,420
And this is basically the
entire reduction, is one line.

1146
01:04:43,420 --> 01:04:48,350
If I have three triples--
if I have a triple xi, xj,

1147
01:04:48,350 --> 01:04:53,530
xk, I'm going to convert
that into a number that

1148
01:04:53,530 --> 01:05:03,150
looks like this where the one
positions are-- I don't really

1149
01:05:03,150 --> 01:05:06,650
know the order, but
they are i, j, and k.

1150
01:05:06,650 --> 01:05:08,190
Everything else is zero.

1151
01:05:08,190 --> 01:05:11,300
And this is in
base b, not base 2.

1152
01:05:11,300 --> 01:05:12,460
It's a little weird.

1153
01:05:12,460 --> 01:05:16,120
All my digits are 0 or
1, but I'm in base b.

1154
01:05:16,120 --> 01:05:18,500
And three of the digits are 1.

1155
01:05:18,500 --> 01:05:19,720
And the rest are zero.

1156
01:05:19,720 --> 01:05:20,440
Why?

1157
01:05:20,440 --> 01:05:22,680
Because of my target sum.

1158
01:05:22,680 --> 01:05:28,470
Target sum is going
to be 1111111111.

1159
01:05:28,470 --> 01:05:31,580
So this number,
in algebra, you're

1160
01:05:31,580 --> 01:05:36,280
write this as b to the i plus
b to the j plus b to the k.

1161
01:05:36,280 --> 01:05:43,437
This you would write as the
sum of b to the i for all i.

1162
01:05:43,437 --> 01:05:44,520
Do you see why this works?

1163
01:05:44,520 --> 01:05:45,992
It's actually really simple.

1164
01:05:51,890 --> 01:05:54,100
For this instance,
my goal is to choose

1165
01:05:54,100 --> 01:05:57,890
a subset of these numbers
that add up to this number.

1166
01:05:57,890 --> 01:05:59,330
How could that possibly happen?

1167
01:05:59,330 --> 01:06:01,810
Well, I've got to choose--
every time I choose

1168
01:06:01,810 --> 01:06:06,930
one of the numbers, those three
digits get set to 1 in my sum.

1169
01:06:06,930 --> 01:06:10,790
If I ever have a collision,
if I add two 1s together,

1170
01:06:10,790 --> 01:06:12,654
I'm going to get a 2.

1171
01:06:12,654 --> 01:06:14,320
That's not good,
because once I get a 2,

1172
01:06:14,320 --> 01:06:16,390
I'll never be able
to get back to a 1,

1173
01:06:16,390 --> 01:06:20,150
because my base is really big.

1174
01:06:20,150 --> 01:06:22,140
This base is designed
so that the total-- this

1175
01:06:22,140 --> 01:06:26,060
is the total number
of colliding 1s.

1176
01:06:26,060 --> 01:06:27,504
So we set it one
larger than that,

1177
01:06:27,504 --> 01:06:29,920
which means you'll never get
a carry when you're adding up

1178
01:06:29,920 --> 01:06:31,000
in this base.

1179
01:06:31,000 --> 01:06:34,420
That's why I set the base to
be something large, not base 2.

1180
01:06:34,420 --> 01:06:39,040
Base 2 might work, but
this is much safer.

1181
01:06:39,040 --> 01:06:44,540
So what that means is for each
of these 1s in the target sum,

1182
01:06:44,540 --> 01:06:47,140
I've got to find a
triple that has those 1s.

1183
01:06:47,140 --> 01:06:48,902
And those triples can't overlap.

1184
01:06:48,902 --> 01:06:51,360
So that means choosing a set
of numbers that add up to this

1185
01:06:51,360 --> 01:06:54,460
is exactly the same as
choosing a set of triples that

1186
01:06:54,460 --> 01:06:56,970
covers all of the elements.

1187
01:06:56,970 --> 01:07:03,150
Done, super easy once you
have the right problem.

1188
01:07:03,150 --> 01:07:05,690
OK, good.

1189
01:07:05,690 --> 01:07:08,750
Now why do I call
this weekly NP-hard?

1190
01:07:08,750 --> 01:07:12,010
Because these numbers are giant.

1191
01:07:12,010 --> 01:07:18,240
If I have n elements
in X, Y, Z over there--

1192
01:07:18,240 --> 01:07:22,130
I guess here they're
called xi, yk, zk.

1193
01:07:22,130 --> 01:07:24,490
Sorry, maybe I should've
called them that here.

1194
01:07:24,490 --> 01:07:26,980
Doesn't matter.

1195
01:07:26,980 --> 01:07:34,120
If I have n of those elements
in X union Y union Z,

1196
01:07:34,120 --> 01:07:38,760
the number of digits here is n.

1197
01:07:38,760 --> 01:07:46,030
So the number of
digits in order n.

1198
01:07:46,030 --> 01:07:48,430
This is fine from an
NP-completeness standpoint.

1199
01:07:48,430 --> 01:07:49,600
This is polynomial size.

1200
01:07:49,600 --> 01:07:55,170
The number of digits in my
numbers is a polynomial.

1201
01:07:55,170 --> 01:07:56,730
And this base is
also pretty small.

1202
01:07:56,730 --> 01:07:58,124
So if you wrote
it out in binary,

1203
01:07:58,124 --> 01:07:59,290
it would also be polynomial.

1204
01:07:59,290 --> 01:08:01,600
So just lost a log factor.

1205
01:08:01,600 --> 01:08:10,460
But the size of the numbers, the
actual values of the numbers,

1206
01:08:10,460 --> 01:08:12,020
is exponential.

1207
01:08:17,350 --> 01:08:20,550
With weak NP-hardness,
that's allowed.

1208
01:08:20,550 --> 01:08:23,920
With strong NP-hardness,
that's forbidden.

1209
01:08:23,920 --> 01:08:26,460
In strong NP-hardness, you
want the values of the numbers

1210
01:08:26,460 --> 01:08:28,370
to be polynomial.

1211
01:08:28,370 --> 01:08:30,950
So in this case, the
number of bits is small,

1212
01:08:30,950 --> 01:08:34,220
but the actual values
are giant, because you

1213
01:08:34,220 --> 01:08:36,271
have to exponentiate.

1214
01:08:36,271 --> 01:08:37,550
It would be cool.

1215
01:08:37,550 --> 01:08:39,279
And this problem is
only weakly NP-hard.

1216
01:08:39,279 --> 01:08:41,700
Maybe you actually know
a pseudo-polynomial time

1217
01:08:41,700 --> 01:08:42,660
algorithm for this.

1218
01:08:42,660 --> 01:08:44,500
It's basically a knapsack.

1219
01:08:44,500 --> 01:08:51,660
If these numbers have polynomial
value, then you can basically,

1220
01:08:51,660 --> 01:08:53,800
in your subproblems in
dynamic programming,

1221
01:08:53,800 --> 01:08:57,609
you can write down
the number t and just

1222
01:08:57,609 --> 01:08:59,149
solve it for all values of t.

1223
01:08:59,149 --> 01:09:01,790
And it's easy to solve
it in polynomial time,

1224
01:09:01,790 --> 01:09:05,302
polynomial in the
integer values.

1225
01:09:05,302 --> 01:09:07,260
So we call that
pseudo-polynomial, because it's

1226
01:09:07,260 --> 01:09:08,229
not really polynomial.

1227
01:09:08,229 --> 01:09:10,065
It's not polynomial in
the number of digits

1228
01:09:10,065 --> 01:09:11,689
that you have to
write down the number.

1229
01:09:11,689 --> 01:09:13,800
It's Polynomial in the values.

1230
01:09:13,800 --> 01:09:18,180
Weak NP-hardness goes together
with pseudo-polynomial.

1231
01:09:18,180 --> 01:09:20,540
That's kind of a matching
result. Say look,

1232
01:09:20,540 --> 01:09:22,890
pseudo-polynomial is
the best you can do.

1233
01:09:22,890 --> 01:09:25,520
You can't hope for a
polynomial because if you

1234
01:09:25,520 --> 01:09:30,330
let the numbers get huge, then
the problem is NP-complete.

1235
01:09:30,330 --> 01:09:34,029
But if you force the numbers to
be small, this problem is easy.

1236
01:09:34,029 --> 01:09:37,010
So subset sum is a little
funny in that sense.

1237
01:09:40,131 --> 01:09:40,630
Cool.

1238
01:10:11,400 --> 01:10:15,327
Let me tell you about
another problem, partition.

1239
01:10:19,360 --> 01:10:21,560
So partition is pretty
much the same set up.

1240
01:10:21,560 --> 01:10:23,200
I'm given n integers.

1241
01:10:31,880 --> 01:10:33,860
Let's say they're positive.

1242
01:10:33,860 --> 01:10:39,040
And I want to know, is
there a subset-- I'm not

1243
01:10:39,040 --> 01:10:41,200
given a target sum t.

1244
01:10:41,200 --> 01:10:42,655
Target sum is basically forced.

1245
01:10:50,240 --> 01:10:53,880
What I would like is the
sum of all the values in S

1246
01:10:53,880 --> 01:10:55,950
to equal the sum
of all the values

1247
01:10:55,950 --> 01:11:00,490
not in S. That's A minus
S, which in other words

1248
01:11:00,490 --> 01:11:04,790
is going to be the sum of
all values in A divided by 2.

1249
01:11:04,790 --> 01:11:08,610
So this is called partition
because you're taking a set,

1250
01:11:08,610 --> 01:11:11,990
you're splitting it into
two halves of equal sum.

1251
01:11:11,990 --> 01:11:14,290
Every element has to go
in one of the two halves.

1252
01:11:14,290 --> 01:11:19,510
And they're called S and A minus
S, like cuts in the flow stuff.

1253
01:11:19,510 --> 01:11:21,227
And you want those
two halves to have

1254
01:11:21,227 --> 01:11:22,810
exactly the same
sum, which means they

1255
01:11:22,810 --> 01:11:24,620
will be the sum divided by 2.

1256
01:11:24,620 --> 01:11:26,320
So that better be
even, otherwise

1257
01:11:26,320 --> 01:11:27,840
it's not going to be possible.

1258
01:11:27,840 --> 01:11:30,190
So again, you want to
decide whether this

1259
01:11:30,190 --> 01:11:34,720
is possible or
impossible, yes or no.

1260
01:11:34,720 --> 01:11:38,200
I claim this problem is
also weakly NP-complete,

1261
01:11:38,200 --> 01:11:42,934
and we can reduce from
subset sum to partition.

1262
01:11:42,934 --> 01:11:45,350
This is a little interesting
because partition is actually

1263
01:11:45,350 --> 01:11:48,560
a special case of subset sum.

1264
01:11:48,560 --> 01:11:53,500
It is the case
where t equals this.

1265
01:11:53,500 --> 01:11:56,490
Subset sum, you're trying to
solve it no matter what t is.

1266
01:11:56,490 --> 01:11:58,280
t is a given input.

1267
01:11:58,280 --> 01:12:00,750
So there's more
instances over here.

1268
01:12:00,750 --> 01:12:03,150
Some of them, some
of these instances

1269
01:12:03,150 --> 01:12:05,650
are the case where t
equals the sum over 2.

1270
01:12:05,650 --> 01:12:07,010
Those are partition instances.

1271
01:12:07,010 --> 01:12:09,550
So this is like a subset
of the possible inputs

1272
01:12:09,550 --> 01:12:11,900
as over there, which
means this problem is

1273
01:12:11,900 --> 01:12:16,470
easier than this one--
no harder anyway.

1274
01:12:16,470 --> 01:12:21,580
In other words, I can reduce
partition to subset sum.

1275
01:12:21,580 --> 01:12:23,750
I just compute this
value and set that to t,

1276
01:12:23,750 --> 01:12:25,880
and then leave the a's alone.

1277
01:12:25,880 --> 01:12:28,350
That will reduce
partition to subset sum.

1278
01:12:28,350 --> 01:12:30,130
But that's not the
direction I want.

1279
01:12:30,130 --> 01:12:32,690
I want to reduce from subset
sum, a problem I can prove

1280
01:12:32,690 --> 01:12:35,360
is NP-complete, to
partition, because I

1281
01:12:35,360 --> 01:12:37,400
want to prove that
partition is NP-complete.

1282
01:12:37,400 --> 01:12:42,090
So in this case, there's an easy
reduction in both directions.

1283
01:12:42,090 --> 01:12:44,750
This direction is
a little harder.

1284
01:12:44,750 --> 01:12:51,410
So reduction from subset sum.

1285
01:12:55,440 --> 01:12:57,700
So I'm given a bunch of ai's.

1286
01:12:57,700 --> 01:12:59,380
I'm not going to touch them.

1287
01:12:59,380 --> 01:13:01,010
And I'm given a target sum t.

1288
01:13:01,010 --> 01:13:04,820
And I basically want to make
that target sum into this half.

1289
01:13:04,820 --> 01:13:09,480
To do that, I'm going to
add two numbers to my set.

1290
01:13:09,480 --> 01:13:17,660
So I'm going to let sigma
be the sum of the given a's.

1291
01:13:17,660 --> 01:13:21,300
And then I'm going to add--
so I'm given a1 through an.

1292
01:13:21,300 --> 01:13:30,720
I'm going to add an plus 1,
is going to be sigma plus t.

1293
01:13:30,720 --> 01:13:40,822
And I'm going to add an plus
2 to be 2 sigma minus t.

1294
01:13:45,050 --> 01:13:45,990
Why?

1295
01:13:45,990 --> 01:13:48,040
So these are two
basically huge numbers.

1296
01:13:51,100 --> 01:13:52,720
Because sigma is
bigger than-- I mean,

1297
01:13:52,720 --> 01:13:54,386
it's the sum of all
the numbers, so it's

1298
01:13:54,386 --> 01:13:56,140
bigger than all of them.

1299
01:13:56,140 --> 01:13:58,320
And so imagine for
a moment that I

1300
01:13:58,320 --> 01:14:02,020
put these two in the same
side of the partition.

1301
01:14:02,020 --> 01:14:04,100
I put them both in S
or I put them both out

1302
01:14:04,100 --> 01:14:09,060
of S. Their sum by
themselves is 3 sigma.

1303
01:14:09,060 --> 01:14:10,450
The t's cancel.

1304
01:14:10,450 --> 01:14:13,940
Whereas all the other
items, their sum is sigma.

1305
01:14:13,940 --> 01:14:15,622
So I'm host.

1306
01:14:15,622 --> 01:14:17,830
If I have 3 sigma on one
side and sigma on the other,

1307
01:14:17,830 --> 01:14:19,830
I'm not going to
make them equal.

1308
01:14:19,830 --> 01:14:23,670
So in fact, these two elements
have to be on opposite sides.

1309
01:14:23,670 --> 01:14:25,690
So there's a side
that has sigma plus t.

1310
01:14:25,690 --> 01:14:28,877
There's a side has
2 sigma minus t.

1311
01:14:28,877 --> 01:14:31,210
And then there's all the other
n items, and some of them

1312
01:14:31,210 --> 01:14:32,990
are going to go to
this side, some of them

1313
01:14:32,990 --> 01:14:35,660
are going to go to this side.

1314
01:14:35,660 --> 01:14:38,310
Their total value is sigma.

1315
01:14:38,310 --> 01:14:40,260
Right now this is
close to sigma.

1316
01:14:40,260 --> 01:14:41,540
This is close to 2 sigma.

1317
01:14:41,540 --> 01:14:43,650
So they have to kind
of meet in the middle.

1318
01:14:43,650 --> 01:14:54,540
In fact, what you'll have to do
is add sigma minus t over here

1319
01:14:54,540 --> 01:15:01,194
and add t over here.

1320
01:15:01,194 --> 01:15:02,360
Think about it for a second.

1321
01:15:02,360 --> 01:15:04,910
If I add sigma minus t,
this comes out to 2 sigma.

1322
01:15:04,910 --> 01:15:07,622
If I add t to this, this
comes out to 2 sigma.

1323
01:15:07,622 --> 01:15:09,330
That would be good
because they're equal.

1324
01:15:09,330 --> 01:15:11,490
And notice that this
is sigma minus t.

1325
01:15:11,490 --> 01:15:12,130
This is t.

1326
01:15:12,130 --> 01:15:13,770
Their sum is sigma.

1327
01:15:13,770 --> 01:15:16,870
So in fact, it has
to be like this.

1328
01:15:16,870 --> 01:15:20,210
You add something over here, and
sigma minus something over here

1329
01:15:20,210 --> 01:15:21,980
for all the other ai's.

1330
01:15:21,980 --> 01:15:24,920
And the something has to be t
in order for these two values

1331
01:15:24,920 --> 01:15:26,390
to equalize.

1332
01:15:26,390 --> 01:15:30,420
So in order to solve this
slightly larger partition

1333
01:15:30,420 --> 01:15:33,640
problem, you have to actually
solve the subset sum problem

1334
01:15:33,640 --> 01:15:37,540
because you have to construct
a subset that adds up to t.

1335
01:15:37,540 --> 01:15:40,260
t was an arbitrary given value.

1336
01:15:40,260 --> 01:15:43,820
So this is pretty nifty.

1337
01:15:43,820 --> 01:15:47,990
We're adding some values so that
the new target sum is the 50/50

1338
01:15:47,990 --> 01:15:51,330
split when we're
given some values that

1339
01:15:51,330 --> 01:15:55,140
have an arbitrary target sum.

1340
01:15:55,140 --> 01:15:59,380
So partition is
weakly NP-complete.

1341
01:15:59,380 --> 01:16:03,220
Let me go to rectangle packing.

1342
01:16:27,690 --> 01:16:30,810
So rectangle packing-- I'm
going to draw a picture.

1343
01:16:30,810 --> 01:16:34,780
I give you a bunch of
rectangles of varying sizes.

1344
01:16:34,780 --> 01:16:37,692
And I give you a
target rectangle.

1345
01:16:37,692 --> 01:16:42,240
Let's call it T.
These are the Ri's.

1346
01:16:42,240 --> 01:16:48,900
I want to put these
rectangles into this picture

1347
01:16:48,900 --> 01:16:52,000
without any overlaps.

1348
01:16:52,000 --> 01:16:54,630
Each of these rectangles
here corresponds to one

1349
01:16:54,630 --> 01:16:56,510
of the rectangles over here.

1350
01:16:56,510 --> 01:17:01,180
So I'll tell you that the sum
of the areas of these rectangles

1351
01:17:01,180 --> 01:17:04,370
is equal to the area of T.
And the question is, can you

1352
01:17:04,370 --> 01:17:08,080
pack those rectangles into
T without any overlaps,

1353
01:17:08,080 --> 01:17:10,850
and therefore without any gaps,
because the areas are exactly

1354
01:17:10,850 --> 01:17:12,090
the same.

1355
01:17:12,090 --> 01:17:16,050
I claim this problem
is weakly NP-hard-- I

1356
01:17:16,050 --> 01:17:21,065
guess NP-complete by
reduction from partition.

1357
01:17:24,740 --> 01:17:31,800
This will be super easy
if you followed what

1358
01:17:31,800 --> 01:17:33,270
the definition of partition is.

1359
01:17:33,270 --> 01:17:36,490
We're given some integers ai.

1360
01:17:36,490 --> 01:17:41,100
And we're going to take each of
them and convert them into a,

1361
01:17:41,100 --> 01:17:46,330
let's say, 1 by 3ai rectangle.

1362
01:17:46,330 --> 01:17:49,490
Three is to avoid some
rotation we'll see.

1363
01:17:49,490 --> 01:17:51,225
And then we're also
given the targets.

1364
01:17:51,225 --> 01:17:53,330
Oh no, target sum is given.

1365
01:17:53,330 --> 01:17:55,210
Target sum is the sum over 2.

1366
01:17:55,210 --> 01:17:59,520
But anyway, we're going to
build our target rectangle

1367
01:17:59,520 --> 01:18:06,970
to be-- it's actually
going to be really big.

1368
01:18:06,970 --> 01:18:13,820
It's going to be 2 by 3 times t.

1369
01:18:13,820 --> 01:18:16,020
So this is that thing.

1370
01:18:16,020 --> 01:18:21,580
So this is 3/2 sum of the a's.

1371
01:18:21,580 --> 01:18:24,720
OK, that's about it.

1372
01:18:24,720 --> 01:18:26,600
In order to pack these
rectangles into here,

1373
01:18:26,600 --> 01:18:28,627
because each of them
is at least three long,

1374
01:18:28,627 --> 01:18:29,960
you cannot pack them vertically.

1375
01:18:29,960 --> 01:18:31,790
They have to be horizontal.

1376
01:18:31,790 --> 01:18:35,320
So in fact what your
packing will look like is

1377
01:18:35,320 --> 01:18:38,070
they'll be the top half
and the bottom half.

1378
01:18:38,070 --> 01:18:40,620
And the top half, the total
length of those rectangles

1379
01:18:40,620 --> 01:18:45,940
has to add up to 3/2 sum of A.
Everything was scaled up by 3,

1380
01:18:45,940 --> 01:18:49,690
so that's 1/2 of A on
the top and the bottom.

1381
01:18:49,690 --> 01:18:50,790
That's a partition.

1382
01:18:50,790 --> 01:18:52,620
In order to pack the
rectangles into here,

1383
01:18:52,620 --> 01:18:55,570
you have to solve the partition
problem, and vice versa.

1384
01:18:55,570 --> 01:18:57,580
Easy.

1385
01:18:57,580 --> 01:19:11,815
OK, let me show you one
more thing, jigsaw puzzles.

1386
01:19:16,790 --> 01:19:21,360
This is not the jigsaw puzzles
you grew up on, somewhat more

1387
01:19:21,360 --> 01:19:23,150
generalized.

1388
01:19:23,150 --> 01:19:27,286
So a piece is going to
look something like this.

1389
01:19:30,480 --> 01:19:33,860
I drew them
intentionally different.

1390
01:19:33,860 --> 01:19:36,820
So on each, you
have a unit square.

1391
01:19:36,820 --> 01:19:39,620
Some of the sides can be flat.

1392
01:19:39,620 --> 01:19:40,840
Some of them can be tabs.

1393
01:19:40,840 --> 01:19:42,500
Some of them can be pockets.

1394
01:19:42,500 --> 01:19:45,000
Each tab and pocket has a shape.

1395
01:19:45,000 --> 01:19:48,110
And they're not in a perfect
matching with each other.

1396
01:19:48,110 --> 01:19:50,950
So there could be
seven of these tabs

1397
01:19:50,950 --> 01:19:53,550
and seven of these pockets,
all the same shape.

1398
01:19:53,550 --> 01:19:55,810
This is what you might call
ambiguous jigsaw puzzles.

1399
01:19:55,810 --> 01:19:58,240
Plus, there is no
image on the piece,

1400
01:19:58,240 --> 01:20:01,900
so this is like
hardcore jigsaw puzzles.

1401
01:20:01,900 --> 01:20:05,100
This is NP-complete.

1402
01:20:05,100 --> 01:20:10,860
And what I'd like to do
is to simulate a rectangle

1403
01:20:10,860 --> 01:20:13,110
with a bunch of jigsaw pieces.

1404
01:20:13,110 --> 01:20:15,790
So it would look
something like this.

1405
01:20:24,080 --> 01:20:27,350
If I have a 1 buy
something rectangle,

1406
01:20:27,350 --> 01:20:31,370
I'm going to simulate it
with that same something,

1407
01:20:31,370 --> 01:20:34,020
little jigsaw pieces.

1408
01:20:34,020 --> 01:20:38,290
And I'm going to make these
shapes only match each other.

1409
01:20:38,290 --> 01:20:41,140
And so for every
rectangle, they're

1410
01:20:41,140 --> 01:20:43,920
going to have a different shape.

1411
01:20:43,920 --> 01:20:45,694
This one will be squares.

1412
01:20:45,694 --> 01:20:47,860
At that point I ran out of
shapes I can easily draw,

1413
01:20:47,860 --> 01:20:48,970
but you get the idea.

1414
01:20:48,970 --> 01:20:50,980
Each rectangle has
a different shape.

1415
01:20:50,980 --> 01:20:52,930
And so these have to
match to each other.

1416
01:20:52,930 --> 01:20:55,000
You can't mix the
tiles, which means you

1417
01:20:55,000 --> 01:20:56,470
have to build this rectangle.

1418
01:20:56,470 --> 01:20:58,300
You have to build
this rectangle.

1419
01:20:58,300 --> 01:21:00,850
And then if the jigsaw
problem is, can you

1420
01:21:00,850 --> 01:21:02,830
fit these into a
given rectangle,

1421
01:21:02,830 --> 01:21:04,460
then you get rectangle packing.

1422
01:21:04,460 --> 01:21:07,185
But this is not a
valid reduction.

1423
01:21:10,610 --> 01:21:15,050
You can't reduce from partition.

1424
01:21:18,550 --> 01:21:20,150
Why?

1425
01:21:20,150 --> 01:21:26,040
Because these numbers are huge.

1426
01:21:26,040 --> 01:21:28,430
Remember, the values of
the numbers in my partition

1427
01:21:28,430 --> 01:21:30,550
instance are exponential.

1428
01:21:30,550 --> 01:21:34,970
So if I have a value ai and it's
exponential in my problem size,

1429
01:21:34,970 --> 01:21:38,030
and I tried to make
ai have little tiles,

1430
01:21:38,030 --> 01:21:40,040
that means a number
of jigsaw pieces

1431
01:21:40,040 --> 01:21:42,814
will be exponential in n.

1432
01:21:42,814 --> 01:21:43,480
That's not good.

1433
01:21:43,480 --> 01:21:45,910
That's not allowed.

1434
01:21:45,910 --> 01:21:49,780
This is why weak
NP-hardness is annoying.

1435
01:21:49,780 --> 01:21:54,550
So instead, we need a
strong NP-hard problem.

1436
01:22:00,020 --> 01:22:01,810
This is a problem
that's NP-hard even when

1437
01:22:01,810 --> 01:22:06,090
the numbers are polynomial
in value, not just in size.

1438
01:22:06,090 --> 01:22:07,400
And it's called 4-partition.

1439
01:22:11,670 --> 01:22:16,970
4-partition, you're given
n integers, as usual.

1440
01:22:20,380 --> 01:22:28,110
Say set is A. And you want
to split those integers

1441
01:22:28,110 --> 01:22:39,990
into n over 4 quadruples
of the same sum.

1442
01:22:46,860 --> 01:22:51,400
So this would be the sum of
A divided by n over four.

1443
01:22:51,400 --> 01:22:53,240
That's your target sum.

1444
01:22:53,240 --> 01:22:55,300
So before we had to
split into two parts that

1445
01:22:55,300 --> 01:22:56,500
had the same sum.

1446
01:22:56,500 --> 01:22:57,600
That was partition.

1447
01:22:57,600 --> 01:23:00,330
Now we have to split
into n over 4 parts.

1448
01:23:00,330 --> 01:23:04,040
Each part will have exactly
four numbers, four integers.

1449
01:23:04,040 --> 01:23:06,840
And they should all
have the same sum.

1450
01:23:06,840 --> 01:23:09,660
This problem is hard
even when the integers

1451
01:23:09,660 --> 01:23:12,345
have polynomial value.

1452
01:23:12,345 --> 01:23:20,765
So the values are at most
some polynomial in n.

1453
01:23:20,765 --> 01:23:22,890
I won't prove it here, but
it's in my lecture notes

1454
01:23:22,890 --> 01:23:23,639
if you're curious.

1455
01:23:23,639 --> 01:23:27,290
It's like this
proof, but harder.

1456
01:23:27,290 --> 01:23:30,090
You end up, instead of
having n digit numbers,

1457
01:23:30,090 --> 01:23:32,080
you have five digit numbers.

1458
01:23:32,080 --> 01:23:36,130
Each digit only has a polynomial
in n different values.

1459
01:23:36,130 --> 01:23:40,470
So the total value of the
numbers is only polynomial.

1460
01:23:40,470 --> 01:23:43,380
It's like n to the
fifth or something.

1461
01:23:43,380 --> 01:23:46,160
Good news is that
this reduction I just

1462
01:23:46,160 --> 01:23:56,710
gave you is also a
reduction from 4-partition

1463
01:23:56,710 --> 01:23:59,950
because it's the same set up.

1464
01:23:59,950 --> 01:24:01,260
Again, I'm given integers.

1465
01:24:01,260 --> 01:24:05,340
Each integer I'm going to
represent by that many tiles.

1466
01:24:05,340 --> 01:24:07,390
Now the number of tiles
is only polynomial,

1467
01:24:07,390 --> 01:24:09,560
so this is a valid reduction.

1468
01:24:09,560 --> 01:24:11,650
And again, if I have to
pack all of these tiles

1469
01:24:11,650 --> 01:24:14,360
into a rectangular board,
that's exactly the same

1470
01:24:14,360 --> 01:24:17,530
as packing these integers.

1471
01:24:17,530 --> 01:24:19,670
Well, I guess I should do
rectangle packing again.

1472
01:24:19,670 --> 01:24:22,590
So this is a proof rectangle
packing was weakly NP-hard.

1473
01:24:22,590 --> 01:24:24,786
But in fact it's
strongly NP-hard.

1474
01:24:24,786 --> 01:24:26,160
You just change
these dimensions.

1475
01:24:26,160 --> 01:24:33,470
You say well, I need whatever,
n over 4 different parts, each

1476
01:24:33,470 --> 01:24:36,820
of size the sum over n over 4.

1477
01:24:36,820 --> 01:24:38,260
You need some scale factor here.

1478
01:24:38,260 --> 01:24:39,360
Three doesn't work.

1479
01:24:39,360 --> 01:24:43,950
Use n or something-- n and n.

1480
01:24:43,950 --> 01:24:46,330
That will prove that
rectangle packing is actually

1481
01:24:46,330 --> 01:24:49,510
strongly NP-hard because
we're reducing for 4-partition

1482
01:24:49,510 --> 01:24:50,612
instead of partition.

1483
01:24:50,612 --> 01:24:52,320
And then you can reduce
rectangle packing

1484
01:24:52,320 --> 01:24:55,187
to jigsaw puzzles because you
have strong hardness over here.

1485
01:24:55,187 --> 01:24:56,520
Over here we don't have numbers.

1486
01:24:56,520 --> 01:24:59,800
We just have these pieces.

1487
01:24:59,800 --> 01:25:01,710
So whenever you convert
from a number problem

1488
01:25:01,710 --> 01:25:05,110
to a non-number problem, if
you're representing the numbers

1489
01:25:05,110 --> 01:25:07,060
in unary, which is
what's going on here,

1490
01:25:07,060 --> 01:25:09,620
you need strong
NP-hardness for it to work.

1491
01:25:09,620 --> 01:25:11,340
Weak NP-hardness isn't enough.

1492
01:25:11,340 --> 01:25:13,970
Then we get jigsaw puzzles,
which we know and love,

1493
01:25:13,970 --> 01:25:15,220
are NP-complete.

1494
01:25:15,220 --> 01:25:17,070
That's it.