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PROFESSOR: OK let's get started.

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00:00:27,790 --> 00:00:31,480
I once taught with
a professor who

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00:00:31,480 --> 00:00:36,100
was lamenting the fact that as
the term progresses attendance

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00:00:36,100 --> 00:00:37,830
in lecture tends to drop off.

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00:00:37,830 --> 00:00:40,640
And gets pretty dramatic
by the end of the term

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00:00:40,640 --> 00:00:43,760
when you're lecturing,
and nobody's there.

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And I asked him what
he did about it.

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00:00:45,980 --> 00:00:48,180
And he thought about
it and he said,

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00:00:48,180 --> 00:00:51,030
there's only two things that
can get students to come

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00:00:51,030 --> 00:00:55,840
to lecture, candy and sex.

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00:00:55,840 --> 00:01:01,510
Now we've already tried
candy, so today we're

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00:01:01,510 --> 00:01:04,410
going to talk about sex.

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00:01:04,410 --> 00:01:07,600
In fact we're going
to use graph theory

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00:01:07,600 --> 00:01:10,630
to address a decades
old debate concerning

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00:01:10,630 --> 00:01:15,240
the relative promiscuity
of men versus women.

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00:01:15,240 --> 00:01:19,410
Now graphs are incredibly useful
structures in computer science,

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00:01:19,410 --> 00:01:22,270
and we're going to be studying
them for the next five or six

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00:01:22,270 --> 00:01:23,520
lectures.

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00:01:23,520 --> 00:01:27,800
They come up in all sorts
of applications, scheduling,

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00:01:27,800 --> 00:01:31,070
optimization,
communications, the design

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00:01:31,070 --> 00:01:32,710
and analysis of algorithms.

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00:01:32,710 --> 00:01:35,310
In fact next week, you're
going to see how to Stanford

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00:01:35,310 --> 00:01:38,990
graduate students
became gazillionaires

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because they use graph
theoretic in a clever way.

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00:01:43,200 --> 00:01:46,470
But let's talk about sex.

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The issue that we're
going to address today

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00:01:48,640 --> 00:01:52,660
is one of the most talked
about, and most well studied,

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00:01:52,660 --> 00:01:55,900
questions in all
of human sociology.

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00:01:55,900 --> 00:02:00,400
On average, who has more
opposite gender partners,

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00:02:00,400 --> 00:02:02,715
men or women?

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00:02:02,715 --> 00:02:05,310
Now opposite gender is
going to be important.

39
00:02:05,310 --> 00:02:09,035
And by this I mean,
one boy, and one girl.

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00:02:09,035 --> 00:02:11,035
All right, I'm not making
a political statement.

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00:02:11,035 --> 00:02:16,770
It's just that the math is a lot
easier that way, as you'll see.

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00:02:16,770 --> 00:02:18,860
Now I'd like to start
by taking a pole here

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00:02:18,860 --> 00:02:21,270
to see what you
think about that.

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00:02:21,270 --> 00:02:24,830
So raise your and if you
think men, on average,

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00:02:24,830 --> 00:02:27,910
have more opposite gender
partners than women do.

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00:02:30,420 --> 00:02:31,095
Only a few.

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00:02:31,095 --> 00:02:33,401
AUDIENCE: In life or [INAUDIBLE]

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PROFESSOR: Um, you can--

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[LAUGHTER]

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00:02:39,040 --> 00:02:40,640
PROFESSOR: One on one.

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00:02:40,640 --> 00:02:43,970
OK, so let's say over the course
of their lives, let's say,

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00:02:43,970 --> 00:02:48,200
or over the course of
2010, that men in America

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00:02:48,200 --> 00:02:51,460
have more opposite-gender
partners than women in America,

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00:02:51,460 --> 00:02:52,830
say in 2010.

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Raise your hand if you think
men have more going on.

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All right a bunch of you.

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00:02:57,040 --> 00:02:58,730
Raise your hand if
you think women have

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more opposite-gender partners?

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00:03:00,890 --> 00:03:02,490
This is unusual.

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Maybe even more voted for
women, but it's close.

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Raise your hand if
you think it's equal.

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All right, about the same.

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Raise your hand if you think
there's no way to know,

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that it's hopeless to
really figure it out.

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All right, nobody goes for that.

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All right, good.

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All right well now in
the popular literature,

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I think the feelings are
different than expressed here.

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Pretty much universally,
in the literature,

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it's believed that men have
more opposite-gender partners

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than women.

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And in fact, you
could even think

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about that, if you
think about literature,

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the leader of the
harem is always a man.

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And he's got lots of women.

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In polygamist
cultures, it's always

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the man that has multiple
wives, not the reverse.

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Now not surprisingly,
this issue has

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been studied
"scientifically," I'll

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00:03:49,620 --> 00:03:54,520
put in quotes, extensively, in
one of the largest studies ever

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done.

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Researchers from
University of Chicago

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interviewed 2,500 people, at
random, over several years.

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They brought them in,
on many occasions,

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to try to get the answer for
the question once and for all.

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And they wrote
this 700 page book,

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called The soul of Social
Organization of Sexuality:

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00:04:14,680 --> 00:04:18,050
Sexual Practices in the US.

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00:04:18,050 --> 00:04:19,670
Actually walking
around with this book

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00:04:19,670 --> 00:04:22,720
has proved to be a
little embarrassing.

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Last week my 11-year-old
daughter saw it,

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and she goes dad, why do
you have this sex book.

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And I grabbed it back and said,
well that's for the course.

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I'm teaching.

95
00:04:34,910 --> 00:04:37,040
And I thought I'd
gotten away with it,

96
00:04:37,040 --> 00:04:38,930
and everything was fine.

97
00:04:38,930 --> 00:04:42,150
And then later that day she
texted all of our friends

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00:04:42,150 --> 00:04:44,330
about the new news
that what do you know,

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00:04:44,330 --> 00:04:48,960
her dad teaches sex ed at MIT.

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00:04:48,960 --> 00:04:53,950
Anyway this study concludes
that on average men

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have 74% more opposite-gender
partners than women.

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There's one other
central claims.

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00:05:10,389 --> 00:05:11,305
And this is in the US.

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OK now, when you think about it
that sounds maybe reasonable,

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00:05:43,350 --> 00:05:45,390
might be OK.

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00:05:45,390 --> 00:05:49,750
But not according to ABC News.

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They did a poll of 1,500
people in the country, in 2004,

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00:05:54,315 --> 00:05:59,090
and concluded that the average
disparity is much greater.

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00:05:59,090 --> 00:06:01,950
In particular, in
this study, they

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00:06:01,950 --> 00:06:05,210
said that the average
man has 20 partners--

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00:06:05,210 --> 00:06:07,100
I'm assuming over
their lifetime--

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00:06:07,100 --> 00:06:10,050
and the average woman has six.

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00:06:10,050 --> 00:06:12,636
And this gives a disparity 233%.

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00:06:21,220 --> 00:06:30,630
So ABC News did a smaller
survey says that it's 233% here,

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00:06:30,630 --> 00:06:33,540
much more than 74%.

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00:06:33,540 --> 00:06:36,800
Now ABC News claimed
this is one of the most

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00:06:36,800 --> 00:06:39,060
scientific studies ever done.

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00:06:39,060 --> 00:06:42,430
And there was a 2.5%
margin of error.

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00:06:42,430 --> 00:06:45,540
Now we'll actually talk about
what that means mathematically

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00:06:45,540 --> 00:06:47,570
later in the term when
we do probability, and do

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00:06:47,570 --> 00:06:49,512
study polling.

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00:06:49,512 --> 00:06:50,970
Now of course I
should also mention

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00:06:50,970 --> 00:06:52,770
that ABC News is
the one that said

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00:06:52,770 --> 00:06:57,720
Al Gore won the presidential
election in 2000.

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00:06:57,720 --> 00:07:01,930
Now the study is called
American Sex Survey,

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00:07:01,930 --> 00:07:04,740
a Peak Between the Sheets.

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00:07:04,740 --> 00:07:06,530
That doesn't sound
so scientific.

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00:07:06,530 --> 00:07:09,990
And it was on TV, on
Primetime Live in 2004.

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00:07:09,990 --> 00:07:12,300
The promo for this
is really good.

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00:07:12,300 --> 00:07:16,740
It says, a groundbreaking ABC
News Primetime Live survey

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00:07:16,740 --> 00:07:20,030
finds a range of eye popping
sexual activities, fantasies,

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00:07:20,030 --> 00:07:22,460
and attitudes in this
country, confirming

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00:07:22,460 --> 00:07:25,700
some conventional wisdom,
exploding some myths,

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00:07:25,700 --> 00:07:28,560
and venturing where few
scientific surveys have

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00:07:28,560 --> 00:07:29,972
gone before.

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By the end of today,
we're going to agree

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00:07:31,680 --> 00:07:34,390
with that last statement.

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00:07:34,390 --> 00:07:39,250
OK now who do you think's right?

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00:07:39,250 --> 00:07:41,980
University of Chicago.

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00:07:41,980 --> 00:07:45,010
Who votes for 74% as
being pretty close?

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A few of you.

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00:07:47,190 --> 00:07:48,890
I've already slammed these guys.

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00:07:48,890 --> 00:07:51,830
Who votes for ABC News
as being more accurate?

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00:07:51,830 --> 00:07:54,980
Yeah, nobody.

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00:07:54,980 --> 00:07:58,210
Who votes for no way to tell?

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00:07:58,210 --> 00:08:00,360
I got some votes
there, all right.

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00:08:00,360 --> 00:08:03,230
So how do you
tackle this problem?

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00:08:03,230 --> 00:08:06,414
In theory we could do
our own 6.042 survey.

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00:08:06,414 --> 00:08:08,080
I don't know how much
we'd really learn,

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00:08:08,080 --> 00:08:10,430
and for sure I'd get fired.

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00:08:10,430 --> 00:08:14,030
So I don't think we're
going to do that.

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00:08:14,030 --> 00:08:16,310
But fortunately, this
is the kind of question

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00:08:16,310 --> 00:08:17,940
that could be
handled, and actually

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00:08:17,940 --> 00:08:21,430
answered, by graph theory,
even though it might

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be more interesting to
interview thousands of people,

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00:08:23,680 --> 00:08:25,640
and find out what's going on.

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00:08:25,640 --> 00:08:28,760
That's not as efficient
as using graphs.

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00:08:28,760 --> 00:08:33,070
So let me start by
defining what a graph is.

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00:08:33,070 --> 00:08:37,210
Informally graph is just
a bunch of dots and lines

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connecting the dots, it's
actually very simple.

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00:08:40,374 --> 00:08:41,165
So here's to graph.

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These are the nodes,
and they're connected

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with these lines, called edges.

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And often the nodes,
and sometimes the edges,

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are labeled.

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For example, we might call this
x1, x2, x3, x4, x5, x6, and x7.

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00:09:10,323 --> 00:09:13,090
So that's an example of a graph.

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00:09:13,090 --> 00:09:14,940
Now this being a
math class, we got

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00:09:14,940 --> 00:09:17,747
to give a formal
definition of a graph.

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00:09:17,747 --> 00:09:19,580
And we'll usually use
the formal definition.

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A graph G is a pair of
sets often called V and E.

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00:09:41,310 --> 00:09:45,850
Where V is a set of elements
called vertices or nodes.

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00:09:45,850 --> 00:09:48,330
And it has to be non-empty
here in this class.

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00:10:06,140 --> 00:10:08,820
And we'll go back and forth
between vertices and nodes.

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00:10:08,820 --> 00:10:13,270
Even the text we use both
words interchangeably.

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00:10:13,270 --> 00:10:26,590
And E is a set of
2-item subsets and V,

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00:10:26,590 --> 00:10:27,950
and they're called edges.

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00:10:32,970 --> 00:10:37,060
So for example, over
here in this picture,

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00:10:37,060 --> 00:10:48,950
V is the set of nodes is x1, x2,
x3, up to x7, that's the nodes.

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00:10:48,950 --> 00:10:54,430
And E, the set of
edges, is pairs,

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00:10:54,430 --> 00:10:58,700
unordered pairs of vertices.

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00:10:58,700 --> 00:11:04,615
So for example
x1, x2 is an edge.

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00:11:04,615 --> 00:11:07,660
And it's the same
as the set x2, x1,

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00:11:07,660 --> 00:11:09,210
doesn't matter the order here.

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00:11:09,210 --> 00:11:11,680
Later in a week or so, we'll
talk about directed graphs

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00:11:11,680 --> 00:11:14,600
where the order matters.

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00:11:14,600 --> 00:11:21,750
x1, x3 is also an
edge here, and so one.

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00:11:21,750 --> 00:11:25,231
Think we've got, let's
see, 1, 2, 3, 4, 5, 6,

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00:11:25,231 --> 00:11:27,570
7 edges in this graph.

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00:11:27,570 --> 00:11:31,250
And the last one
would be x5, x7.

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00:11:35,300 --> 00:11:41,190
Edges are also sometimes written
with this notation, x1 line x2,

192
00:11:41,190 --> 00:11:42,877
is another notation.

193
00:11:42,877 --> 00:11:44,960
And then later when you
talk about directed edges,

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00:11:44,960 --> 00:11:49,710
we'll put a little arrowhead
on one end of this.

195
00:11:49,710 --> 00:11:52,110
Now the definition of a graph
is really pretty simple.

196
00:11:52,110 --> 00:11:55,290
Just think of it as dots
and lines, if you want.

197
00:11:55,290 --> 00:11:59,440
But there's often differences
in how people define graphs.

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00:11:59,440 --> 00:12:04,550
For example, in this class we
don't allow the empty graph,

199
00:12:04,550 --> 00:12:06,742
i.e. the graph with no nodes.

200
00:12:06,742 --> 00:12:08,950
So we're going to insist
that every graph has to have

201
00:12:08,950 --> 00:12:10,270
at least one node in it.

202
00:12:10,270 --> 00:12:11,770
And that's just to
make the theorems

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00:12:11,770 --> 00:12:13,010
we're going prove be true.

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00:12:13,010 --> 00:12:14,510
Otherwise there's
some theorems that

205
00:12:14,510 --> 00:12:16,880
are false for the special
case of the empty graph.

206
00:12:16,880 --> 00:12:19,270
But we don't require the
graph to have any edges.

207
00:12:19,270 --> 00:12:23,330
In fact, it's possible you
have a graph with nodes,

208
00:12:23,330 --> 00:12:23,945
but no edges.

209
00:12:26,500 --> 00:12:28,560
For example, this graph.

210
00:12:28,560 --> 00:12:29,310
Three-node graph.

211
00:12:33,560 --> 00:12:42,730
So here G equals VE,
V equals x1, x2, x3.

212
00:12:42,730 --> 00:12:44,210
And E is just the empty set.

213
00:12:48,110 --> 00:12:53,060
Now for a general graph,
when you do have edges,

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00:12:53,060 --> 00:13:04,005
we say that two nodes,
call them xi and xj,

215
00:13:04,005 --> 00:13:13,120
are adjacent if they're
connected by an edge, namely

216
00:13:13,120 --> 00:13:18,160
if xi xj is an edge.

217
00:13:18,160 --> 00:13:24,630
All right so for example,
x5 is adjacent to x7,

218
00:13:24,630 --> 00:13:29,040
but it's not adjacent to
x4, there's no edge there.

219
00:13:33,160 --> 00:13:37,810
Closely related is the
definition of the incidence.

220
00:13:37,810 --> 00:13:48,410
An edge E, which
is xixj, is said

221
00:13:48,410 --> 00:13:55,560
to be incident to its
end points, xi and xj.

222
00:13:59,120 --> 00:14:06,400
OK so, for example, if I
labeled that edge as E, E

223
00:14:06,400 --> 00:14:11,435
is the edge x1, x2, and this
incident to x1, and incident

224
00:14:11,435 --> 00:14:11,935
x2.

225
00:14:18,320 --> 00:14:20,600
Then we can talk about
the degree of a node.

226
00:14:23,410 --> 00:14:38,180
The number of edges
incident to a node

227
00:14:38,180 --> 00:14:40,030
is called the
degree of the node.

228
00:14:51,510 --> 00:14:57,520
So for example, what's the
degree of x5 over here?

229
00:14:57,520 --> 00:15:05,010
3, so in this case, the
degree of x5 equals 3.

230
00:15:05,010 --> 00:15:10,920
The degree of x7 is 1.

231
00:15:10,920 --> 00:15:14,190
These guys all have
degree 0, there's

232
00:15:14,190 --> 00:15:17,980
no edges incident to them.

233
00:15:17,980 --> 00:15:21,790
Now in this class, we're going
to look at only simple graphs,

234
00:15:21,790 --> 00:15:23,000
at lease for a while.

235
00:15:26,570 --> 00:15:40,950
A graph is simple if it has
no loops, or multiple edges.

236
00:15:48,200 --> 00:15:53,750
Now a loop is an edge that only
connects up one node, that's

237
00:15:53,750 --> 00:15:56,250
a loop and we don't allow it.

238
00:15:56,250 --> 00:16:00,070
A multiple edge is we've got two
edges that are really the same,

239
00:16:00,070 --> 00:16:02,980
they connect the same endpoints.

240
00:16:02,980 --> 00:16:04,040
Also called a multi-edge.

241
00:16:08,290 --> 00:16:10,620
And those we're not going
to have in simple graphs.

242
00:16:10,620 --> 00:16:11,975
We don't allow this.

243
00:16:11,975 --> 00:16:15,200
We don't allow that.

244
00:16:15,200 --> 00:16:17,920
Any questions so far
about what a graph is?

245
00:16:22,490 --> 00:16:26,390
So how are we going
to use a graph

246
00:16:26,390 --> 00:16:32,470
to model the problem of
opposite-gender partners?

247
00:16:32,470 --> 00:16:34,890
That's the question we're after.

248
00:16:34,890 --> 00:16:38,052
So any thoughts about what
the nodes of the graph

249
00:16:38,052 --> 00:16:39,010
are going to represent?

250
00:16:41,840 --> 00:16:42,440
What is it?

251
00:16:42,440 --> 00:16:43,930
AUDIENCE: Males and females?

252
00:16:43,930 --> 00:16:45,370
PROFESSOR: People.

253
00:16:45,370 --> 00:16:47,670
Yeah, so we're going
to have people.

254
00:16:47,670 --> 00:16:50,290
In fact, there's two
kinds of people here.

255
00:16:50,290 --> 00:16:52,630
There's men, and women.

256
00:16:55,370 --> 00:17:01,520
All right we got nodes
here for the men.

257
00:17:01,520 --> 00:17:05,400
And in fact in America,
there's a lot of nodes here.

258
00:17:05,400 --> 00:17:07,750
All right, and so this
might be oh I don't know,

259
00:17:07,750 --> 00:17:14,310
say that's Tom Cruise
and Nicole Kidman.

260
00:17:14,310 --> 00:17:16,944
Now what's the edge
going to represent?

261
00:17:16,944 --> 00:17:18,420
AUDIENCE: Partners.

262
00:17:18,420 --> 00:17:20,020
PROFESSOR: Partners.

263
00:17:20,020 --> 00:17:23,109
They were
opposite-gender partners.

264
00:17:23,109 --> 00:17:26,470
And there's actually
more edges probably here.

265
00:17:26,470 --> 00:17:34,200
We could have Penelope
here, and Katie here.

266
00:17:34,200 --> 00:17:37,400
And well probably lots more, I
probably don't know them all.

267
00:17:37,400 --> 00:17:41,250
And Ben's over here with Nicole.

268
00:17:41,250 --> 00:17:48,054
And Nicole got Jude and Keith.

269
00:17:48,054 --> 00:17:49,470
There's actually
a website you can

270
00:17:49,470 --> 00:17:53,500
go to get a lot of
these things here.

271
00:17:53,500 --> 00:17:56,830
And Katie went with Josh.

272
00:17:56,830 --> 00:17:59,779
It's called whosedatedwho.com,
and you get big graph,

273
00:17:59,779 --> 00:18:01,320
you could start
filling in the edges.

274
00:18:01,320 --> 00:18:03,690
I don't know how reliable it is.

275
00:18:03,690 --> 00:18:06,340
Now it's really critical
that we're only looking edges

276
00:18:06,340 --> 00:18:08,270
from here to here.

277
00:18:08,270 --> 00:18:10,650
All right, so if there's an
edge between Tom and Ben,

278
00:18:10,650 --> 00:18:13,740
I don't want to know about it.

279
00:18:13,740 --> 00:18:17,540
Just opposite-gender partners.

280
00:18:17,540 --> 00:18:22,090
OK now in the USA,
the number of nodes

281
00:18:22,090 --> 00:18:25,590
here is about 300 million.

282
00:18:28,640 --> 00:18:30,220
About three million people.

283
00:18:30,220 --> 00:18:35,080
And the number of men nodes,
male nodes, call these VM,

284
00:18:35,080 --> 00:18:39,950
and this is VW, by the way,
I'm using cardinality notation.

285
00:18:39,950 --> 00:18:42,750
When I put bars
around a set, that

286
00:18:42,750 --> 00:18:46,050
is the denoting how
many are in the set.

287
00:18:46,050 --> 00:18:53,360
In the US there's about
147.6 men out of the 300.

288
00:18:53,360 --> 00:18:58,820
And the number of women--
oh we got a w here--

289
00:18:58,820 --> 00:19:03,240
is about 152.4 million.

290
00:19:03,240 --> 00:19:05,740
So there's a little
bit more nodes

291
00:19:05,740 --> 00:19:08,420
on this side of the graph,
than that side in the US.

292
00:19:10,960 --> 00:19:12,130
What about the edges?

293
00:19:12,130 --> 00:19:14,115
Any idea of how many
edges there are here?

294
00:19:17,097 --> 00:19:17,680
We don't know.

295
00:19:17,680 --> 00:19:21,070
I sure as heck don't know
how many edges there are.

296
00:19:21,070 --> 00:19:22,410
So that we don't know.

297
00:19:22,410 --> 00:19:26,160
The cardinality of the
edge set we don't know,

298
00:19:26,160 --> 00:19:27,940
and we're not likely
to figure out.

299
00:19:27,940 --> 00:19:30,420
I don't even think
these surveys, really,

300
00:19:30,420 --> 00:19:32,900
can estimate that.

301
00:19:32,900 --> 00:19:35,430
But what we're
trying to figure out

302
00:19:35,430 --> 00:19:40,570
is the ratio of the
average degree of the men,

303
00:19:40,570 --> 00:19:44,310
to the average
degree of the women.

304
00:19:44,310 --> 00:19:46,280
Because the number of
opposite-gender partners

305
00:19:46,280 --> 00:19:48,640
you have is your
degree here, and you're

306
00:19:48,640 --> 00:19:50,770
looking for the
average guy degree,

307
00:19:50,770 --> 00:19:54,370
compared to the average
female degree here.

308
00:19:54,370 --> 00:19:55,390
That's what we're after.

309
00:19:55,390 --> 00:19:57,835
All right so let's
find that quantity.

310
00:20:00,740 --> 00:20:10,020
Let's let A sub m equal
the average number

311
00:20:10,020 --> 00:20:21,240
of opposite-gender
partners for men.

312
00:20:24,830 --> 00:20:27,970
And we can let A W be
the same thing for women.

313
00:20:37,310 --> 00:20:38,090
All right.

314
00:20:38,090 --> 00:20:42,440
Now we're trying to figure out
the answer to this question.

315
00:20:42,440 --> 00:20:48,360
What is A m, the
average guy degree,

316
00:20:48,360 --> 00:20:52,550
over the average woman degree.

317
00:20:52,550 --> 00:20:54,490
And in particular, the
University of Chicago

318
00:20:54,490 --> 00:20:58,480
says, they say it's 1.74.

319
00:20:58,480 --> 00:21:02,770
That the average guy as 74%
more opposite-gender partners

320
00:21:02,770 --> 00:21:04,550
than the average woman.

321
00:21:04,550 --> 00:21:10,830
ABC News says it's 3.33, that
is 233% more for the men,

322
00:21:10,830 --> 00:21:12,460
than the women.

323
00:21:12,460 --> 00:21:16,860
Now we're going to figure
this out what this ratio is.

324
00:21:16,860 --> 00:21:18,540
Just use a little
bit of math here,

325
00:21:18,540 --> 00:21:21,260
and a little bit
of graph theory.

326
00:21:21,260 --> 00:21:23,650
So let's write a
formula for A m.

327
00:21:27,760 --> 00:21:34,230
Well we're trying to figure out
the average degree over here.

328
00:21:34,230 --> 00:21:35,660
Well, that's pretty simple.

329
00:21:35,660 --> 00:21:39,370
We just add up all the
degrees, and divide

330
00:21:39,370 --> 00:21:40,735
by the number of nodes.

331
00:21:40,735 --> 00:21:42,360
And that'll give us
the average degree.

332
00:21:44,910 --> 00:21:48,390
So the average degree is
the sum of the degrees,

333
00:21:48,390 --> 00:21:51,670
over all men, x
in the set of men,

334
00:21:51,670 --> 00:21:56,490
of the degree of x, divided
by the number of men.

335
00:21:59,830 --> 00:22:04,620
Can somebody give me a
simpler expression for this?

336
00:22:04,620 --> 00:22:07,454
It doesn't have that
nasty sum in it?

337
00:22:07,454 --> 00:22:09,210
AUDIENCE: E.

338
00:22:09,210 --> 00:22:12,440
PROFESSOR: E. The
cardinality of E.

339
00:22:12,440 --> 00:22:16,960
I'm adding all the degrees here.

340
00:22:16,960 --> 00:22:20,940
Well that's just another way
of counting all the edges,

341
00:22:20,940 --> 00:22:24,430
because every edge shows
up once, and only once,

342
00:22:24,430 --> 00:22:26,360
in a degree count here.

343
00:22:26,360 --> 00:22:28,990
And this is where,
we use the fact we

344
00:22:28,990 --> 00:22:30,370
have opposite-gender partners.

345
00:22:33,020 --> 00:22:36,130
Because if I had
some edges over here

346
00:22:36,130 --> 00:22:40,240
they wouldn't get counted
in sum of the degrees here.

347
00:22:40,240 --> 00:22:45,255
All right so this is
just the cardinality

348
00:22:45,255 --> 00:22:48,970
of the number of edges,
divided by the number of men.

349
00:22:48,970 --> 00:22:50,380
Any questions about that?

350
00:22:50,380 --> 00:22:53,020
Because this is an
important statement about

351
00:22:53,020 --> 00:22:54,930
graphs in general.

352
00:22:54,930 --> 00:22:57,384
When I have a graph
like this-- which

353
00:22:57,384 --> 00:22:58,800
is called a bipartite
graph, we'll

354
00:22:58,800 --> 00:23:00,249
talk about more in a little bit.

355
00:23:00,249 --> 00:23:02,290
But where the edges go
from the left to the right

356
00:23:02,290 --> 00:23:03,987
if I sum the
degrees on the left,

357
00:23:03,987 --> 00:23:05,570
I'm just counting
the number of edges.

358
00:23:09,060 --> 00:23:13,840
All right, let's figure out a
formula for the average number

359
00:23:13,840 --> 00:23:16,340
of partners for the women.

360
00:23:16,340 --> 00:23:23,240
That simple that's just
sum x over the women.

361
00:23:23,240 --> 00:23:27,950
The degree of x, divided
by the number of women.

362
00:23:30,900 --> 00:23:32,530
Let me rewrite that
so it's clearer.

363
00:23:36,150 --> 00:23:38,508
What's a simpler
expression for this?

364
00:23:38,508 --> 00:23:39,383
AUDIENCE: [INAUDIBLE]

365
00:23:43,080 --> 00:23:46,310
PROFESSOR: Yeah,
this sum, adding

366
00:23:46,310 --> 00:23:52,410
the degrees of the women,
is just the number of edges,

367
00:23:52,410 --> 00:23:54,690
right.

368
00:23:54,690 --> 00:24:00,280
So that is cardinality of edges,
divided by the number of women.

369
00:24:00,280 --> 00:24:06,700
All right, well now we can
write, solve for our formula,

370
00:24:06,700 --> 00:24:12,320
average over men over
average of the women.

371
00:24:12,320 --> 00:24:22,126
That's E over VM,
divided by E over VW.

372
00:24:22,126 --> 00:24:24,900
Wow, this is nice.

373
00:24:24,900 --> 00:24:30,290
I don't know the number of edges
is, but it just canceled out.

374
00:24:30,290 --> 00:24:33,300
And this is just
the number of women,

375
00:24:33,300 --> 00:24:34,890
divided by the number of men.

376
00:24:37,430 --> 00:24:40,740
And in fact we know that.

377
00:24:40,740 --> 00:24:42,660
That's this number,
divided by that number,

378
00:24:42,660 --> 00:24:52,090
which is about 1.0325.

379
00:24:52,090 --> 00:24:57,550
So we just proved,
that on average, a man

380
00:24:57,550 --> 00:25:01,950
has 3%, or 3 and 1/4% more
opposite-gender partners

381
00:25:01,950 --> 00:25:02,500
than women.

382
00:25:05,010 --> 00:25:08,800
No need to do the interviews,
or spend years doing.

383
00:25:08,800 --> 00:25:10,390
That is the answer.

384
00:25:10,390 --> 00:25:14,120
And it has nothing to do
with the promiscuity of men,

385
00:25:14,120 --> 00:25:16,345
or women, nothing at all.

386
00:25:19,080 --> 00:25:23,300
So the Chicago study is way
off, and the ABC New study

387
00:25:23,300 --> 00:25:25,400
is completely nuts.

388
00:25:25,400 --> 00:25:30,050
It just can't be
right, this is a proof.

389
00:25:30,050 --> 00:25:33,740
Now what happened here?

390
00:25:33,740 --> 00:25:37,910
Well what's going on, what's
the reason for why this is true?

391
00:25:37,910 --> 00:25:38,410
Yeah?

392
00:25:38,410 --> 00:25:41,679
AUDIENCE: A male has
a female partner then

393
00:25:41,679 --> 00:25:44,004
the female has a male partner.

394
00:25:44,004 --> 00:25:44,670
PROFESSOR: Yeah.

395
00:25:44,670 --> 00:25:47,456
AUDIENCE: You're not looking
at like how many males

396
00:25:47,456 --> 00:25:49,786
are going to one female.

397
00:25:49,786 --> 00:25:53,110
The promiscuity isn't even
a part of the question.

398
00:25:53,110 --> 00:25:54,110
PROFESSOR: That's right.

399
00:25:54,110 --> 00:25:56,610
It takes two to tango.

400
00:25:56,610 --> 00:25:58,880
Every time you got a
guy, you got a women.

401
00:25:58,880 --> 00:26:01,920
And you have the number
of relationships going.

402
00:26:01,920 --> 00:26:04,530
The average for the men is that
number, divided by the men.

403
00:26:04,530 --> 00:26:06,810
Average for the women
is that same number,

404
00:26:06,810 --> 00:26:07,699
divided by the women.

405
00:26:07,699 --> 00:26:09,240
And so if there's
more women, they're

406
00:26:09,240 --> 00:26:11,080
going to have less
partners on average.

407
00:26:11,080 --> 00:26:12,670
Has to be.

408
00:26:12,670 --> 00:26:14,420
So it really was
a stupid question.

409
00:26:14,420 --> 00:26:17,970
It's very, very
simple to answer.

410
00:26:17,970 --> 00:26:21,340
Now as it turns out there are
endless studies like this,

411
00:26:21,340 --> 00:26:23,090
in the literature.

412
00:26:23,090 --> 00:26:25,940
In fact, a few years
ago the Boston Globe

413
00:26:25,940 --> 00:26:31,140
ran an explosive story about
the study habits of students

414
00:26:31,140 --> 00:26:33,530
on Boston-area campuses.

415
00:26:33,530 --> 00:26:38,320
And their surveys show that,
on average, minority students

416
00:26:38,320 --> 00:26:41,380
tended to study with
non-minority students

417
00:26:41,380 --> 00:26:42,685
more than the other way around.

418
00:26:46,350 --> 00:26:49,250
And they want on great
length consulting the experts

419
00:26:49,250 --> 00:26:50,980
as to why this might be true.

420
00:26:50,980 --> 00:26:53,230
Why is it the minority
students study

421
00:26:53,230 --> 00:26:56,870
with non-minority students
more than the other way around.

422
00:26:56,870 --> 00:26:58,940
Now can anyone tell
me why it is certainly

423
00:26:58,940 --> 00:27:02,205
true, and not surprising,
why that's the case?

424
00:27:02,205 --> 00:27:04,120
AUDIENCE: Because
they're the minority.

425
00:27:04,120 --> 00:27:05,703
PROFESSOR: Because
they're a minority.

426
00:27:05,703 --> 00:27:07,745
There's fewer minorities
than non-minorities.

427
00:27:07,745 --> 00:27:12,030
End of story, we don't
need this sociology PhD

428
00:27:12,030 --> 00:27:13,780
from down the street
to explain it to us.

429
00:27:16,780 --> 00:27:19,060
We're going to see a lot of
other bogus studies later.

430
00:27:19,060 --> 00:27:22,270
This is not unusual, especially
when we get the probability.

431
00:27:22,270 --> 00:27:26,410
Just every day there's a
new one in probability.

432
00:27:26,410 --> 00:27:29,310
Any questions about
this before we leave?

433
00:27:29,310 --> 00:27:32,160
Unfortunately that's most all
we'll say about sex today.

434
00:27:35,590 --> 00:27:36,800
OK.

435
00:27:36,800 --> 00:27:42,110
But now, in this example,
we used an edge in the graph

436
00:27:42,110 --> 00:27:46,530
to denote some kind of
affinity between two nodes.

437
00:27:46,530 --> 00:27:48,594
The two nodes liked
each other in some sense

438
00:27:48,594 --> 00:27:50,010
if they were
connected by an edge,

439
00:27:50,010 --> 00:27:52,520
or they had a
relationship of some kind.

440
00:27:52,520 --> 00:27:54,490
There's lots of examples
in computer science

441
00:27:54,490 --> 00:27:57,453
where you use an edge to
denote just the opposite.

442
00:27:57,453 --> 00:28:00,200
That the two nodes can't
be near each other,

443
00:28:00,200 --> 00:28:02,750
or don't like each other.

444
00:28:02,750 --> 00:28:04,420
For example,
consider the problem

445
00:28:04,420 --> 00:28:08,080
of scheduling
final exams at MIT.

446
00:28:08,080 --> 00:28:11,310
And they do this after they
find out all of your schedules,

447
00:28:11,310 --> 00:28:14,354
and they try to schedule
the exams so that you don't

448
00:28:14,354 --> 00:28:16,270
have to take two at once,
or there's as little

449
00:28:16,270 --> 00:28:19,270
of that as possible.

450
00:28:19,270 --> 00:28:22,730
For example, let's
do an example here.

451
00:28:33,460 --> 00:28:35,760
Say we look at
these five classes.

452
00:28:44,590 --> 00:28:47,220
Take 6041.

453
00:28:47,220 --> 00:28:49,920
And this may not be totally
accurate, but roughly.

454
00:28:57,250 --> 00:28:59,950
So I've got five
MIT classes, and I'm

455
00:28:59,950 --> 00:29:03,920
going to put an edge between
pairs of classes that have

456
00:29:03,920 --> 00:29:05,400
overlapping student enrollment.

457
00:29:08,630 --> 00:29:11,490
So in this case,
for example, we've

458
00:29:11,490 --> 00:29:13,160
assumed in the
drawing of his graph,

459
00:29:13,160 --> 00:29:17,280
that you can't have our
exam the same time is 6002,

460
00:29:17,280 --> 00:29:20,640
on the assumption there's
students in both classes.

461
00:29:20,640 --> 00:29:24,560
But you could have our
exam the same time as 6034.

462
00:29:24,560 --> 00:29:27,730
Because there's not
an overlapping student

463
00:29:27,730 --> 00:29:29,350
in both classes,
so the exams could

464
00:29:29,350 --> 00:29:31,510
be scheduled at the same time.

465
00:29:31,510 --> 00:29:33,720
So we've used a
graph to represent

466
00:29:33,720 --> 00:29:38,740
which courses can't have
their exam at the same time.

467
00:29:38,740 --> 00:29:42,100
Now let's also suppose we have
a set of slots for the exam.

468
00:29:45,360 --> 00:29:47,670
And say they're
all on a Wednesday.

469
00:29:47,670 --> 00:29:51,480
And the first slot is
Wednesday from 5:00 to 7:00.

470
00:29:51,480 --> 00:29:55,430
And the next one
is 7:00 to 9:00.

471
00:29:55,430 --> 00:30:00,070
And then, the next
one is 9:00 to 11:00.

472
00:30:00,070 --> 00:30:05,290
And then 11:00 to 1:00 in the
morning, and then 1:00 to 3:00,

473
00:30:05,290 --> 00:30:07,930
getting pretty late.

474
00:30:07,930 --> 00:30:13,230
And your job is to figure
out how not to have

475
00:30:13,230 --> 00:30:15,290
to use these later exam slots.

476
00:30:15,290 --> 00:30:17,140
You'd like to use
as few as possible

477
00:30:17,140 --> 00:30:20,180
so you're not going
too late night,

478
00:30:20,180 --> 00:30:22,930
or come before the
holidays, so you're not

479
00:30:22,930 --> 00:30:25,380
having exams on Christmas
and New Year's, for example.

480
00:30:28,450 --> 00:30:33,840
So the goal is to assign
slots to the nodes.

481
00:30:33,840 --> 00:30:37,090
Put every node in
a slot so you don't

482
00:30:37,090 --> 00:30:42,110
have nodes hooked by an
edge getting the same slot.

483
00:30:42,110 --> 00:30:44,400
Now this is an example
of what's called

484
00:30:44,400 --> 00:30:46,496
a graph coloring problem.

485
00:30:46,496 --> 00:30:47,370
So let's define that.

486
00:31:17,680 --> 00:31:44,020
Given a graph G, and K colors,
assign a color to each node,

487
00:31:44,020 --> 00:31:55,970
so that adjacent nodes
get different colors.

488
00:32:03,154 --> 00:32:08,120
All right, and then the minimum
number of colors you need

489
00:32:08,120 --> 00:32:11,750
is called the chromatic
number of the graph.

490
00:32:11,750 --> 00:32:36,460
So the minimum value of K, for
which such a coloring exist,

491
00:32:36,460 --> 00:32:48,840
is the chromatic
number OF the graph.

492
00:32:48,840 --> 00:32:59,956
And it's denoted by
this symbol chi of G.

493
00:32:59,956 --> 00:33:02,330
Because usually you want to
use a small number of colors.

494
00:33:02,330 --> 00:33:07,270
Now what does a color
represent when we're

495
00:33:07,270 --> 00:33:09,220
dealing with this problem?

496
00:33:09,220 --> 00:33:10,760
What's the meaning of a color?

497
00:33:10,760 --> 00:33:11,900
AUDIENCE: Time slot.

498
00:33:11,900 --> 00:33:14,080
PROFESSOR: A time slot, OK.

499
00:33:14,080 --> 00:33:22,690
So let's call this time slot
C1, C2, C3, C4, C5, so there's

500
00:33:22,690 --> 00:33:23,850
five possible colors.

501
00:33:23,850 --> 00:33:26,800
Now of course, we could color
this graph with five colors,

502
00:33:26,800 --> 00:33:29,150
every node could just
get its own color.

503
00:33:29,150 --> 00:33:33,060
But then somebody's taking
their exam from 1:00 to 3:00 AM,

504
00:33:33,060 --> 00:33:36,880
and that's a bit of a pain.

505
00:33:36,880 --> 00:33:41,360
Let's see if we can
do less than five.

506
00:33:41,360 --> 00:33:46,010
Let's say I give this color
one, let's give this one color

507
00:33:46,010 --> 00:33:50,630
one, that's OK, because
they're not connected.

508
00:33:50,630 --> 00:33:55,880
I can't give this one color one,
so I give it color two, say.

509
00:33:55,880 --> 00:33:58,570
Now this one I can't give color
one, because this guy got it,

510
00:33:58,570 --> 00:34:00,570
he can't get color two,
because that guy got it.

511
00:34:00,570 --> 00:34:03,020
So it give it color three.

512
00:34:03,020 --> 00:34:08,159
And well, I can't do
one, two, or three here,

513
00:34:08,159 --> 00:34:11,020
so I gotta go to color four.

514
00:34:11,020 --> 00:34:17,800
All right so 6042 will get
the 11:00 PM to 1:00 AM slot,

515
00:34:17,800 --> 00:34:19,443
not so good.

516
00:34:19,443 --> 00:34:21,916
Can we do any better?

517
00:34:21,916 --> 00:34:26,139
Can we get away
with three colors.

518
00:34:26,139 --> 00:34:29,870
Some say yes, some say no.

519
00:34:29,870 --> 00:34:33,690
How many people think you can
do three colors on this graph?

520
00:34:33,690 --> 00:34:34,830
A bunch.

521
00:34:34,830 --> 00:34:37,402
How many think you
can't do any better?

522
00:34:37,402 --> 00:34:40,270
All right, the vote
is mostly for three.

523
00:34:40,270 --> 00:34:42,060
Let's see.

524
00:34:42,060 --> 00:34:42,639
Any ideas?

525
00:34:42,639 --> 00:34:43,980
Anybody see how to do three?

526
00:34:46,775 --> 00:34:47,275
Yeah?

527
00:34:47,275 --> 00:34:50,199
AUDIENCE: Assign C4 to 6034 .

528
00:34:50,199 --> 00:34:52,550
PROFESSOR: Assign C4 to 6043.

529
00:34:52,550 --> 00:34:55,889
AUDIENCE: Or C1 to 6042.

530
00:34:55,889 --> 00:34:59,580
PROFESSOR: C-- I can't
do see C1 to 6042.

531
00:34:59,580 --> 00:35:01,561
It crashes, but can I do-- yeah?

532
00:35:01,561 --> 00:35:02,060
Put

533
00:35:02,060 --> 00:35:04,434
AUDIENCE: C1 in 6003.

534
00:35:04,434 --> 00:35:05,350
PROFESSOR: C1 in 6003.

535
00:35:05,350 --> 00:35:10,222
AUDIENCE: And get
rid of C1 in 6034.

536
00:35:10,222 --> 00:35:11,180
PROFESSOR: Get rid of--

537
00:35:11,180 --> 00:35:12,580
AUDIENCE: Make it C2.

538
00:35:12,580 --> 00:35:15,070
PROFESSOR: Make this a C2.

539
00:35:15,070 --> 00:35:17,362
Oh, yeah.

540
00:35:17,362 --> 00:35:21,520
All right, these got C1,
they're not adjacent.

541
00:35:21,520 --> 00:35:23,740
These got C2,
they're not adjacent.

542
00:35:23,740 --> 00:35:26,380
This can now get C3.

543
00:35:26,380 --> 00:35:29,990
So we can have our exam from
9:00 to 11:00, which is better.

544
00:35:29,990 --> 00:35:32,375
All right, can anybody
do it in two colors?

545
00:35:40,780 --> 00:35:44,935
Can anybody offer a reason why
two colors may not be possible?

546
00:35:47,536 --> 00:35:48,035
Yeah?

547
00:35:48,035 --> 00:35:51,575
AUDIENCE: Because let's say you
could do it with two colors.

548
00:35:51,575 --> 00:35:52,200
PROFESSOR: Yep.

549
00:35:52,200 --> 00:35:58,170
AUDIENCE: 6041 and 6002
have to be different colors.

550
00:35:58,170 --> 00:35:59,830
PROFESSOR: Yes.

551
00:35:59,830 --> 00:36:03,500
AUDIENCE: 6042 can't be
C1, and it can't be C2.

552
00:36:03,500 --> 00:36:05,300
PROFESSOR: Yeah, good.

553
00:36:05,300 --> 00:36:09,130
So you can't in two colors,
because these three guys

554
00:36:09,130 --> 00:36:10,670
would violate that.

555
00:36:10,670 --> 00:36:11,990
You've got a triangle here.

556
00:36:11,990 --> 00:36:15,550
Each one of these guys has to
be different than the other two.

557
00:36:15,550 --> 00:36:16,920
So two colors can't work.

558
00:36:16,920 --> 00:36:19,740
You've got to have at
least three in this case.

559
00:36:19,740 --> 00:36:20,760
So three is optimal.

560
00:36:20,760 --> 00:36:24,000
We have just shown
for this graph,

561
00:36:24,000 --> 00:36:26,836
the chromatic number is three.

562
00:36:26,836 --> 00:36:33,300
All right, now in general doing
what we just did is very hard.

563
00:36:33,300 --> 00:36:35,680
No one knows a fast
algorithm for determining

564
00:36:35,680 --> 00:36:38,560
the chromatic number.

565
00:36:38,560 --> 00:36:40,370
In fact, it's a weird
kind of problem,

566
00:36:40,370 --> 00:36:44,239
because it's easy enough to
check that a coloring is OK.

567
00:36:44,239 --> 00:36:46,530
If somebody put a coloring
on the board, you can check,

568
00:36:46,530 --> 00:36:48,420
oh that works really simply.

569
00:36:48,420 --> 00:36:52,070
Just check every edge, and make
sure the colors are different.

570
00:36:52,070 --> 00:36:54,590
But figuring it out,
as best we know,

571
00:36:54,590 --> 00:36:57,840
you've got to try an exponential
number of possibilities.

572
00:36:57,840 --> 00:37:00,962
So if I had 100 nodes
here, my running time

573
00:37:00,962 --> 00:37:02,920
of the algorithm to check
all the possibilities

574
00:37:02,920 --> 00:37:04,800
would be exponential
and a hundred.

575
00:37:04,800 --> 00:37:05,300
Yeah?

576
00:37:05,300 --> 00:37:08,772
AUDIENCE: Can that number
just like the highest

577
00:37:08,772 --> 00:37:11,750
degree of each node, or nodes.

578
00:37:11,750 --> 00:37:13,015
PROFESSOR: Uh no.

579
00:37:13,015 --> 00:37:15,770
But it's no worse than
something like that,

580
00:37:15,770 --> 00:37:17,130
as we'll see a few minutes.

581
00:37:17,130 --> 00:37:18,750
That's a great observation.

582
00:37:18,750 --> 00:37:21,210
And we're going to come back
to that in a few minutes.

583
00:37:21,210 --> 00:37:24,180
But it's not just that.

584
00:37:24,180 --> 00:37:30,140
OK now in fact even figuring
out for an arbitrary graph

585
00:37:30,140 --> 00:37:33,320
if three colors
can be done, called

586
00:37:33,320 --> 00:37:36,290
the three-coloring problem,
that's really hard.

587
00:37:36,290 --> 00:37:39,570
No one knows how to solve that
in less than exponential time.

588
00:37:39,570 --> 00:37:43,000
In fact, one of these
NP-complete problems

589
00:37:43,000 --> 00:37:44,340
is what it's called.

590
00:37:44,340 --> 00:37:47,870
How many people here don't
know about NP-completeness?

591
00:37:47,870 --> 00:37:49,822
Is everybody-- all
right so all of you

592
00:37:49,822 --> 00:37:51,030
haven't seen NP-completeness.

593
00:37:51,030 --> 00:37:56,120
OK so there is a class of
thousands of problems--

594
00:37:56,120 --> 00:38:00,230
in fact there's books list these
1,000 problems-- that are all

595
00:38:00,230 --> 00:38:03,000
NP-complete, somebody's proved
they belong in the class.

596
00:38:03,000 --> 00:38:06,440
And what that means is
that if somebody gave you

597
00:38:06,440 --> 00:38:09,300
a solution, like
a coloring here,

598
00:38:09,300 --> 00:38:13,850
it's easy to check really
quickly if it's valid.

599
00:38:13,850 --> 00:38:16,040
But figuring it
out is really hard.

600
00:38:16,040 --> 00:38:17,790
And if you figured
out how to solve

601
00:38:17,790 --> 00:38:19,761
one of those
thousands of problems,

602
00:38:19,761 --> 00:38:22,260
like suddenly you figured out
how to tell if any graph could

603
00:38:22,260 --> 00:38:25,610
work with three colors, you
would solve automatically

604
00:38:25,610 --> 00:38:28,429
all other thousands in the book.

605
00:38:28,429 --> 00:38:30,470
So it's this book of
problems you will constantly

606
00:38:30,470 --> 00:38:33,952
run into in your career
in computer science.

607
00:38:33,952 --> 00:38:35,410
And it's bad when
you run into one,

608
00:38:35,410 --> 00:38:38,200
because there's no good
algorithm to solve it known.

609
00:38:38,200 --> 00:38:40,065
But if you just
solved one of them,

610
00:38:40,065 --> 00:38:43,620
the other thousands would
suddenly be solvable quickly.

611
00:38:43,620 --> 00:38:46,540
Even better, you win a
million dollar prize.

612
00:38:46,540 --> 00:38:48,000
One of these
Millennium Prizes we

613
00:38:48,000 --> 00:38:50,860
talked about the first lecture.

614
00:38:50,860 --> 00:38:53,880
Even if you show you can't
find a fast algorithm for one

615
00:38:53,880 --> 00:38:57,000
of them, that means that known
of them have fast algorithms,

616
00:38:57,000 --> 00:38:59,397
and you also get
a million dollars.

617
00:38:59,397 --> 00:39:01,980
So this is the central problem
in computer science, and theory

618
00:39:01,980 --> 00:39:04,410
computing, is whether
or not you could solve

619
00:39:04,410 --> 00:39:07,060
these NP-complete problems.

620
00:39:07,060 --> 00:39:09,770
Now actually lots of
people have claim to do it.

621
00:39:09,770 --> 00:39:12,300
And in fact, there was a
lot of buzz in the community

622
00:39:12,300 --> 00:39:15,110
about a month ago when
actually a reputable researcher

623
00:39:15,110 --> 00:39:17,370
at HP Labs said he'd done it.

624
00:39:17,370 --> 00:39:21,630
He proved that you can't
solve NP-complete problems.

625
00:39:21,630 --> 00:39:24,100
And he got people going for
probably at least a week,

626
00:39:24,100 --> 00:39:26,420
until they discovered
a fatal flaw.

627
00:39:26,420 --> 00:39:29,550
And the proof was
actually bogus.

628
00:39:29,550 --> 00:39:31,770
So no one still knows
if you can solve

629
00:39:31,770 --> 00:39:34,770
these NP-complete
problems quickly.

630
00:39:34,770 --> 00:39:38,620
Now the problem is, in practice,
you run into these things

631
00:39:38,620 --> 00:39:40,430
all the time, like
MIT really does

632
00:39:40,430 --> 00:39:42,430
have to schedule the exams.

633
00:39:42,430 --> 00:39:43,680
So you've got to do something.

634
00:39:43,680 --> 00:39:46,690
You can't just go say, hey
it's NP-complete, so no exams

635
00:39:46,690 --> 00:39:48,870
this year, or whatever.

636
00:39:48,870 --> 00:39:52,970
That's not going to fly,
so you got to do something.

637
00:39:52,970 --> 00:39:55,320
So now this is a
problem-- many of you

638
00:39:55,320 --> 00:39:58,510
when you go into careers, you're
going to be faced with this.

639
00:39:58,510 --> 00:40:00,250
You got to do something.

640
00:40:00,250 --> 00:40:06,010
Any thoughts about an algorithm
for coloring graphs that might

641
00:40:06,010 --> 00:40:07,880
use a small number of colors?

642
00:40:07,880 --> 00:40:10,335
It doesn't have to
always work, or you're

643
00:40:10,335 --> 00:40:11,960
going to win a lot
of money if it does.

644
00:40:11,960 --> 00:40:14,300
But a simple
algorithm, you can't

645
00:40:14,300 --> 00:40:16,590
take either the 100 steps.

646
00:40:16,590 --> 00:40:19,690
You got to be linear,
probably, or quadratic time.

647
00:40:19,690 --> 00:40:22,800
That could get you a
small number of colors.

648
00:40:22,800 --> 00:40:25,860
Any thoughts about
what you'd do?

649
00:40:25,860 --> 00:40:26,360
Yeah?

650
00:40:26,360 --> 00:40:28,770
AUDIENCE: The number
of degrees and nodes?

651
00:40:28,770 --> 00:40:30,590
PROFESSOR: The
number-- what about it?

652
00:40:30,590 --> 00:40:33,220
AUDIENCE: The highest
degree and that node,

653
00:40:33,220 --> 00:40:34,799
the 6042 is [INAUDIBLE].

654
00:40:34,799 --> 00:40:35,465
PROFESSOR: Yeah.

655
00:40:35,465 --> 00:40:38,700
AUDIENCE: So you could use that.

656
00:40:38,700 --> 00:40:40,210
PROFESSOR: Good, all right.

657
00:40:40,210 --> 00:40:42,200
So what do I do with
that-- so I found

658
00:40:42,200 --> 00:40:45,130
a node with a high
degree, there's

659
00:40:45,130 --> 00:40:47,030
three of them have
degree three here.

660
00:40:47,030 --> 00:40:48,806
What do I do with them?

661
00:40:48,806 --> 00:40:52,324
AUDIENCE: Pick a
different color to.

662
00:40:52,324 --> 00:40:53,740
PROFESSOR: Pick a
different color,

663
00:40:53,740 --> 00:40:55,890
that means I've colored
some of the others.

664
00:40:55,890 --> 00:40:59,040
If I pick a different
color, do I start with them,

665
00:40:59,040 --> 00:41:03,030
or do I finish with
a high degree nodes?

666
00:41:03,030 --> 00:41:05,807
Because you've got to
assign the colors to them.

667
00:41:05,807 --> 00:41:07,890
And high degree is important
to be thinking about.

668
00:41:07,890 --> 00:41:10,431
We're going to prove a theorem
in just a minute about related

669
00:41:10,431 --> 00:41:11,730
to degree and coloring.

670
00:41:11,730 --> 00:41:14,800
AUDIENCE: Start with them.

671
00:41:14,800 --> 00:41:17,310
PROFESSOR: Start with
them, and do what with it?

672
00:41:17,310 --> 00:41:18,498
Color?

673
00:41:18,498 --> 00:41:22,890
AUDIENCE: Yeah, and then assign
the ones that aren't connected

674
00:41:22,890 --> 00:41:26,310
[INAUDIBLE] to the same slots.

675
00:41:26,310 --> 00:41:29,250
PROFESSOR: OK, so I could--
here's a degree of theory

676
00:41:29,250 --> 00:41:32,630
now I can start with
color one for that.

677
00:41:32,630 --> 00:41:34,190
And then what do I do next?

678
00:41:34,190 --> 00:41:38,372
I pick-- its neighbors have to
get different colors, I guess.

679
00:41:38,372 --> 00:41:39,830
You'd start coloring
the neighbors.

680
00:41:39,830 --> 00:41:42,305
AUDIENCE: My first
instinct would be

681
00:41:42,305 --> 00:41:44,780
to color all the [INAUDIBLE].

682
00:41:48,250 --> 00:41:49,530
PROFESSOR: OK.

683
00:41:49,530 --> 00:41:51,320
And what color
would use for them?

684
00:41:51,320 --> 00:41:52,702
AUDIENCE: Different ones.

685
00:41:52,702 --> 00:41:54,660
PROFESSOR: Different ones
if they're connected,

686
00:41:54,660 --> 00:41:57,790
or if they're not connected
you'd still use different ones?

687
00:41:57,790 --> 00:42:00,670
AUDIENCE: Only if
they're connected.

688
00:42:00,670 --> 00:42:02,940
PROFESSOR: Only they're
connected use different ones.

689
00:42:02,940 --> 00:42:06,901
And so if they're not connected,
you'd use the same colors?

690
00:42:06,901 --> 00:42:07,401
Yeah?

691
00:42:10,190 --> 00:42:14,620
You're going close, and it
actually works pretty well.

692
00:42:14,620 --> 00:42:17,910
The underlying principle you're
sort of thinking about here

693
00:42:17,910 --> 00:42:21,160
is you've got some notion
of the order in which you're

694
00:42:21,160 --> 00:42:22,985
going to process your graph.

695
00:42:22,985 --> 00:42:25,110
And you're going to start
with a high degree nodes,

696
00:42:25,110 --> 00:42:26,530
in your case.

697
00:42:26,530 --> 00:42:28,200
And as you go
along, you're going

698
00:42:28,200 --> 00:42:30,950
to start coloring the nodes.

699
00:42:30,950 --> 00:42:33,990
And you're going to make
sure you color them legally.

700
00:42:33,990 --> 00:42:37,920
And it sounds like you're going
to color them with a low color

701
00:42:37,920 --> 00:42:39,270
as you go along.

702
00:42:39,270 --> 00:42:44,860
And that is probably the most
basic graph coloring approach.

703
00:42:44,860 --> 00:42:48,560
And almost you could almost
say is a generic approach.

704
00:42:48,560 --> 00:42:52,091
So let's define that, and then
see prove some facts about it.

705
00:42:56,710 --> 00:42:58,860
Most of the graph coloring
algorithms in practice

706
00:42:58,860 --> 00:43:01,070
are based on this approach.

707
00:43:01,070 --> 00:43:05,320
And we're going to call it the
basic graph coloring algorithm.

708
00:43:14,240 --> 00:43:20,840
And for our graph G, with
vertices V, and edges E.

709
00:43:20,840 --> 00:43:31,080
So the first step is going to be
to order the nodes from 1 to n.

710
00:43:34,110 --> 00:43:35,850
Now in your case,
you were suggesting

711
00:43:35,850 --> 00:43:39,700
an ordering where I have
the high degree nodes first.

712
00:43:39,700 --> 00:43:41,217
All right.

713
00:43:41,217 --> 00:43:43,050
But for now we're not
going to specify that.

714
00:43:43,050 --> 00:43:47,290
We're going to make it
any ordering you want.

715
00:43:47,290 --> 00:43:50,080
And then we're going to
have a notion of an order

716
00:43:50,080 --> 00:43:51,490
on the colors, as well.

717
00:43:57,815 --> 00:43:59,190
And I don't know
how many colors,

718
00:43:59,190 --> 00:44:02,480
but they're going to be
numbered 1, 2, and so forth.

719
00:44:02,480 --> 00:44:08,773
And then we're going to process
the nodes one at a time,

720
00:44:08,773 --> 00:44:13,860
to N. We color the
nodes, what is step I,

721
00:44:13,860 --> 00:44:18,430
we color the Ith node V sub i
with the lowest legal color.

722
00:44:25,240 --> 00:44:28,490
And by the legal I mean you
don't color at the same node

723
00:44:28,490 --> 00:44:31,640
as another node that's already
been colored the same that it's

724
00:44:31,640 --> 00:44:32,602
adjacent to.

725
00:44:40,790 --> 00:44:43,290
All right so let's try this.

726
00:44:43,290 --> 00:44:47,480
In fact, this is sort of the
algorithm I used initially

727
00:44:47,480 --> 00:44:50,700
to color exam graph over there.

728
00:44:54,508 --> 00:44:58,230
All right, so
let's look at that.

729
00:44:58,230 --> 00:45:04,310
So let's say we-- let me
erase the colors here, and put

730
00:45:04,310 --> 00:45:05,465
an ordering on the nodes.

731
00:45:08,100 --> 00:45:12,230
So let's say I ordered
them with 6034 first,

732
00:45:12,230 --> 00:45:15,060
so this would be V1.

733
00:45:15,060 --> 00:45:18,390
Then 6041 is V2.

734
00:45:18,390 --> 00:45:23,380
Then V3, V4, V5.

735
00:45:26,290 --> 00:45:30,552
If that's my ordering, what
color would I assign to 6034?

736
00:45:30,552 --> 00:45:31,730
AUDIENCE: One.

737
00:45:31,730 --> 00:45:35,140
PROFESSOR: One, C1, I'd
color it first to get C1.

738
00:45:35,140 --> 00:45:38,860
What color does 6041 get?

739
00:45:38,860 --> 00:45:41,420
C1, as well, it's the
lowest possible color

740
00:45:41,420 --> 00:45:47,110
that's legal, and is not hooked
to this guy, so C1 is legal.

741
00:45:47,110 --> 00:45:49,810
What color do I give here?

742
00:45:49,810 --> 00:45:52,480
C2.

743
00:45:52,480 --> 00:45:59,170
Then I color this one next
C-- can't do C2, can't do C1,

744
00:45:59,170 --> 00:46:01,460
so I pick C3.

745
00:46:01,460 --> 00:46:03,500
And then I get to
6042 last, and I

746
00:46:03,500 --> 00:46:07,220
can't do one, two, or
three, so I do four.

747
00:46:07,220 --> 00:46:09,150
All right so algorithm,
with that ordering,

748
00:46:09,150 --> 00:46:10,860
gave four colors.

749
00:46:10,860 --> 00:46:14,970
However we know there's a way
to do a different ordering that

750
00:46:14,970 --> 00:46:19,220
gives us three colors.

751
00:46:19,220 --> 00:46:20,840
In particular,
let's see if we do

752
00:46:20,840 --> 00:46:23,150
this what happens if we
use this other ordering.

753
00:46:23,150 --> 00:46:26,050
Let me erase these.

754
00:46:26,050 --> 00:46:33,960
Say that's V1, V2, V3, V4, V5.

755
00:46:33,960 --> 00:46:40,140
Now I get C1, this
will be C2, C1.

756
00:46:43,920 --> 00:46:45,670
What's this one get?

757
00:46:45,670 --> 00:46:46,320
C2.

758
00:46:46,320 --> 00:46:49,350
Ah, much better.

759
00:46:49,350 --> 00:46:50,800
C3.

760
00:46:50,800 --> 00:46:54,880
So different orderings result
in different numbers of colors

761
00:46:54,880 --> 00:46:56,380
here.

762
00:46:56,380 --> 00:47:00,850
So the whole art now becomes
finding a clever ordering.

763
00:47:00,850 --> 00:47:03,210
And so many people have
already had good ideas,

764
00:47:03,210 --> 00:47:05,850
pick the largest
degree nodes first.

765
00:47:05,850 --> 00:47:11,740
And in fact, if you simulate
the algorithm on lots of graphs,

766
00:47:11,740 --> 00:47:16,140
you do better on average when
you color the larger degree

767
00:47:16,140 --> 00:47:17,820
nodes first.

768
00:47:17,820 --> 00:47:20,850
And then if you start to
use more exotic orderings,

769
00:47:20,850 --> 00:47:22,762
you can do even better.

770
00:47:22,762 --> 00:47:24,720
If you take a lot of
graphs that are out there,

771
00:47:24,720 --> 00:47:26,719
and run your algorithm,
and see how well you do,

772
00:47:26,719 --> 00:47:29,540
you do better with more
sophisticated orderings.

773
00:47:29,540 --> 00:47:34,550
In fact, this was my
senior thesis back

774
00:47:34,550 --> 00:47:36,290
when I was
undergraduate student.

775
00:47:36,290 --> 00:47:39,650
I was trying to figure out
better and better orderings

776
00:47:39,650 --> 00:47:41,690
that worked for graphs.

777
00:47:41,690 --> 00:47:44,820
And at the time it caused
a bit of a problem.

778
00:47:44,820 --> 00:47:47,470
I was a undergraduate
at Princeton.

779
00:47:47,470 --> 00:47:49,120
And Princeton, to
this day I think,

780
00:47:49,120 --> 00:47:52,090
still has exams after the
holidays, the Christmas

781
00:47:52,090 --> 00:47:54,010
holidays, New Year's holidays.

782
00:47:54,010 --> 00:47:57,810
And the students wanted to have
the exams before Christmas,

783
00:47:57,810 --> 00:47:59,770
because they hated going
home for the holiday,

784
00:47:59,770 --> 00:48:01,790
and then you've got to
worry about your exams

785
00:48:01,790 --> 00:48:03,270
when you come back.

786
00:48:03,270 --> 00:48:06,050
And the faculty said
no, there's no way

787
00:48:06,050 --> 00:48:09,570
to get them all compressed
into a small number of days.

788
00:48:09,570 --> 00:48:12,010
Now I wasn't aware of
all that of the time.

789
00:48:12,010 --> 00:48:14,255
But my thesis was go
figure out good ordering.

790
00:48:14,255 --> 00:48:15,880
So I tried lots of
different orderings.

791
00:48:15,880 --> 00:48:17,850
And I tried the
largest degree first,

792
00:48:17,850 --> 00:48:21,730
and recursive versions of that
actually worked very well.

793
00:48:21,730 --> 00:48:25,620
And then tried it on the
Princeton exam graph.

794
00:48:25,620 --> 00:48:28,310
And lo and behold, you could
actually squish it down,

795
00:48:28,310 --> 00:48:30,200
so you could give all
the exams, I think was,

796
00:48:30,200 --> 00:48:33,960
4 and 1/2 days, plenty of time
to give them before Christmas.

797
00:48:33,960 --> 00:48:36,330
Which caused a fair of
scandal at the time,

798
00:48:36,330 --> 00:48:38,670
because then the faculty
had to come clean

799
00:48:38,670 --> 00:48:40,730
that they just didn't
want to bother having

800
00:48:40,730 --> 00:48:44,180
the exams before Christmas.

801
00:48:44,180 --> 00:48:46,770
Now this algorithm is
an example of what's

802
00:48:46,770 --> 00:48:48,780
known as a greedy algorithm.

803
00:48:48,780 --> 00:48:52,930
Now in a greedy algorithm
it's always simple.

804
00:48:52,930 --> 00:48:55,360
You just go one
step after the next,

805
00:48:55,360 --> 00:48:57,760
taking the best you
can do at each stop.

806
00:48:57,760 --> 00:49:00,670
You never go back and try
to make things better.

807
00:49:00,670 --> 00:49:03,950
You never do hill climbing, if
you're familiar with that term.

808
00:49:03,950 --> 00:49:07,010
You just always keep it simple,
one thing after the next,

809
00:49:07,010 --> 00:49:08,830
very fast.

810
00:49:08,830 --> 00:49:10,770
Sometimes it works
great in practice.

811
00:49:10,770 --> 00:49:12,570
Sometimes it doesn't.

812
00:49:12,570 --> 00:49:16,860
But it's always where you start,
some simple approach like this.

813
00:49:16,860 --> 00:49:20,770
Now this algorithm
actually, even

814
00:49:20,770 --> 00:49:22,850
if you don't try to
monkey with the ordering,

815
00:49:22,850 --> 00:49:25,140
even for a worst case
ordering of the nodes,

816
00:49:25,140 --> 00:49:28,930
that actually does pretty
good for a lot of graphs.

817
00:49:28,930 --> 00:49:30,730
And in fact, it
does really well--

818
00:49:30,730 --> 00:49:33,040
as somebody already
asked about--

819
00:49:33,040 --> 00:49:35,510
if all the nodes
have low degree.

820
00:49:35,510 --> 00:49:37,060
So let's state
that as a theorem.

821
00:49:37,060 --> 00:49:38,518
And then we're
going to prove that.

822
00:49:44,329 --> 00:49:54,330
So if every node in a graph
G has degree, at most,

823
00:49:54,330 --> 00:49:58,550
d-- so that's the biggest
degree in the graph, D-- then

824
00:49:58,550 --> 00:50:16,480
this basic algorithm uses, at
most, d plus 1 colors for G.

825
00:50:16,480 --> 00:50:19,920
No matter what the
ordering is, you'll

826
00:50:19,920 --> 00:50:23,490
never do worse than
d plus 1 colors.

827
00:50:23,490 --> 00:50:28,840
So what's the value of d for
our exam graph over here?

828
00:50:28,840 --> 00:50:30,180
d is 3.

829
00:50:30,180 --> 00:50:33,240
Every node has degree,
at most, three.

830
00:50:33,240 --> 00:50:37,120
And so it says, that no matter
what ordering you picked here,

831
00:50:37,120 --> 00:50:39,350
you'd get at most four colors.

832
00:50:39,350 --> 00:50:40,350
Now you might do better.

833
00:50:40,350 --> 00:50:42,620
In fact, we found an
ordering that got three.

834
00:50:42,620 --> 00:50:45,940
So it's possible to do better.

835
00:50:45,940 --> 00:50:49,040
So let's prove this fact
because this makes a difference.

836
00:50:49,040 --> 00:50:53,260
Say you have a graph
with hundreds of nodes.

837
00:50:53,260 --> 00:50:55,716
But every node has
degree, at most, three.

838
00:50:55,716 --> 00:50:58,560
Well that says you only
need four colors even,

839
00:50:58,560 --> 00:51:01,840
if the graph has 1,000 nodes,
and that's very useful.

840
00:51:01,840 --> 00:51:04,580
So in that kind of
situation it does very well.

841
00:51:04,580 --> 00:51:05,525
So let's prove that.

842
00:51:09,570 --> 00:51:12,733
Any ideas as to what proof
technique we're going to use?

843
00:51:12,733 --> 00:51:14,705
AUDIENCE: Invariant.

844
00:51:14,705 --> 00:51:17,120
PROFESSOR: Invariant, close.

845
00:51:17,120 --> 00:51:19,560
Not quite an
invariant, but close.

846
00:51:19,560 --> 00:51:22,450
AUDIENCE: [INAUDIBLE]

847
00:51:22,450 --> 00:51:23,404
PROFESSOR: What?

848
00:51:23,404 --> 00:51:24,820
AUDIENCE: Well
ordering principle.

849
00:51:24,820 --> 00:51:26,420
PROFESSOR: You know well
ordering principle, yeah,

850
00:51:26,420 --> 00:51:28,503
we're going to use the
equivalent version of that.

851
00:51:28,503 --> 00:51:31,024
We're going to use induction.

852
00:51:31,024 --> 00:51:33,190
If you like well-- it's
equivalent to well ordering.

853
00:51:33,190 --> 00:51:35,314
If you like well ordering
you could do it that way.

854
00:51:35,314 --> 00:51:38,400
I think it's easier
using induction here.

855
00:51:38,400 --> 00:51:39,765
So the proof is by induction.

856
00:51:45,110 --> 00:51:46,720
All right so the
first thing we need

857
00:51:46,720 --> 00:51:47,845
is an induction hypothesis.

858
00:51:50,950 --> 00:51:54,450
Any thoughts about what the
induction hypothesis should be?

859
00:51:54,450 --> 00:51:55,202
Yeah?

860
00:51:55,202 --> 00:52:01,270
AUDIENCE: If you have a
graph with n nodes then

861
00:52:01,270 --> 00:52:06,150
where the degree of any nodes
is less than [INAUDIBLE]

862
00:52:06,150 --> 00:52:08,090
then you can do it.

863
00:52:08,090 --> 00:52:09,090
PROFESSOR: That's great.

864
00:52:09,090 --> 00:52:12,030
You're going to do really
well on the midterm,

865
00:52:12,030 --> 00:52:14,450
because you put an
n into this thing,

866
00:52:14,450 --> 00:52:16,570
but there's not an
n here to start.

867
00:52:16,570 --> 00:52:18,320
What are most
people going to do--

868
00:52:18,320 --> 00:52:20,210
we used to ask this actually.

869
00:52:20,210 --> 00:52:22,830
We asked this once on
a test many years ago,

870
00:52:22,830 --> 00:52:25,820
and it was an utter disaster,
because did everybody do?

871
00:52:25,820 --> 00:52:28,832
May be one student, or
two, put an n into there.

872
00:52:28,832 --> 00:52:31,290
But what's the naturally thing
to do to induct on here when

873
00:52:31,290 --> 00:52:34,560
you look at this statement?

874
00:52:34,560 --> 00:52:37,760
You're going to induct on d,
because the first thing you do

875
00:52:37,760 --> 00:52:39,680
is you make this be your
induction hypothesis.

876
00:52:39,680 --> 00:52:41,950
There's only one thing
to use, so you're

877
00:52:41,950 --> 00:52:44,910
going to have your
predicate be p of d,

878
00:52:44,910 --> 00:52:47,014
and it's going to be that.

879
00:52:47,014 --> 00:52:49,680
Now It didn't occur to us that's
what everybody was going to do,

880
00:52:49,680 --> 00:52:50,560
but it should have.

881
00:52:50,560 --> 00:52:53,220
They all did that and
it was a disaster.

882
00:52:53,220 --> 00:52:56,010
Because if you do
this, well you've

883
00:52:56,010 --> 00:53:00,280
got to take a graph with
maximum degree d, or d plus 1

884
00:53:00,280 --> 00:53:03,200
in the inductive step,
pull out all the nodes

885
00:53:03,200 --> 00:53:07,021
with degree d plus 1 to get
a graph with now degree d.

886
00:53:07,021 --> 00:53:07,770
And that's a mess.

887
00:53:07,770 --> 00:53:10,410
You just pulled out a lot
of nodes, potentially.

888
00:53:10,410 --> 00:53:14,650
Color that in d plus 1 colors,
now put all that junk back in.

889
00:53:14,650 --> 00:53:17,180
And say only used
one more color.

890
00:53:17,180 --> 00:53:18,520
Nightmare.

891
00:53:18,520 --> 00:53:21,300
And these were MIT
students under pressure.

892
00:53:21,300 --> 00:53:23,330
It was a nightmare.

893
00:53:23,330 --> 00:53:24,410
So that does not work.

894
00:53:24,410 --> 00:53:27,410
And in fact, we will ask
an induction question

895
00:53:27,410 --> 00:53:29,960
on graphs on every test
you take in this course.

896
00:53:29,960 --> 00:53:31,190
It will happen.

897
00:53:31,190 --> 00:53:34,850
And so usually,
with induction, you

898
00:53:34,850 --> 00:53:36,580
take this as your
induction hypothesis.

899
00:53:36,580 --> 00:53:38,650
With graphs, you
have to be careful.

900
00:53:38,650 --> 00:53:41,830
And worst part about
this is we tell people

901
00:53:41,830 --> 00:53:45,950
when this doesn't work, use a
stronger induction hypothesis.

902
00:53:45,950 --> 00:53:48,280
So students tried
to make a stronger,

903
00:53:48,280 --> 00:53:51,900
but they're still stuck on d,
and it was still a disaster.

904
00:53:51,900 --> 00:53:53,930
With graphs, you do
something different.

905
00:53:53,930 --> 00:53:56,240
And the first thing you
do with a graph, usually,

906
00:53:56,240 --> 00:53:58,385
is put n in here.

907
00:53:58,385 --> 00:54:00,510
And if it doesn't work with
n, the number of nodes,

908
00:54:00,510 --> 00:54:03,410
you put in e the
number of edges.

909
00:54:03,410 --> 00:54:05,120
And induct on that.

910
00:54:05,120 --> 00:54:08,460
And so what you said is
exactly the right thing to do.

911
00:54:08,460 --> 00:54:11,270
Don't do this, or least don't
spend too much time on it.

912
00:54:11,270 --> 00:54:13,360
Pretty quickly try this.

913
00:54:13,360 --> 00:54:17,970
If every end node graph--
if every node in an n

914
00:54:17,970 --> 00:54:23,840
node graph G has
degree at most degree,

915
00:54:23,840 --> 00:54:27,070
then the basic algorithm uses
at most d, plus one colors.

916
00:54:27,070 --> 00:54:30,350
And now you induct on n.

917
00:54:30,350 --> 00:54:33,100
And almost always on graphs,
that's the first thing to try.

918
00:54:33,100 --> 00:54:37,270
Even if it's not in
your theorem statement.

919
00:54:37,270 --> 00:54:39,900
Any questions about that?

920
00:54:46,940 --> 00:54:49,570
Well let's start
with this, and see

921
00:54:49,570 --> 00:54:51,256
if we can make this one work.

922
00:55:02,230 --> 00:55:04,140
So what's the next
step in our proof?

923
00:55:04,140 --> 00:55:06,560
What do we got to do?

924
00:55:06,560 --> 00:55:07,380
Base case.

925
00:55:11,890 --> 00:55:14,750
And the base case will
be, not n equals 0,

926
00:55:14,750 --> 00:55:19,300
because we can't have a zero
node graph, but n equals 1.

927
00:55:19,300 --> 00:55:22,940
And how many edges do we have?

928
00:55:22,940 --> 00:55:23,440
Zero.

929
00:55:23,440 --> 00:55:25,190
If there's one node,
we don't allow loops,

930
00:55:25,190 --> 00:55:33,140
so it's zero edges, which means
that the degree of our graph

931
00:55:33,140 --> 00:55:36,380
has to be zero.

932
00:55:36,380 --> 00:55:37,780
There's no edges.

933
00:55:37,780 --> 00:55:39,900
And of course there's
only one node,

934
00:55:39,900 --> 00:55:44,140
so one color is going
to work, and that

935
00:55:44,140 --> 00:55:46,340
happens to equal d plus 1.

936
00:55:49,460 --> 00:55:52,110
All right, so the
base case is true.

937
00:55:52,110 --> 00:55:55,110
For one node graphs,
you can always

938
00:55:55,110 --> 00:55:59,010
use d plus 1 colors,
where d is the max degree.

939
00:55:59,010 --> 00:56:01,210
All right, next we have
the inductive step.

940
00:56:05,500 --> 00:56:14,425
So here we assume P n is
true for the induction.

941
00:56:20,330 --> 00:56:23,130
And now we look at an
n plus 1 node graph

942
00:56:23,130 --> 00:56:26,620
to show P n plus 1 is true.

943
00:56:26,620 --> 00:56:35,240
So we let G be any
N plus 1 node graph.

944
00:56:35,240 --> 00:56:39,152
We got to show you can
color it in d plus 1 colors.

945
00:56:41,810 --> 00:56:48,790
And let's let d be the
max degree, the largest

946
00:56:48,790 --> 00:56:56,020
degree in G.

947
00:56:56,020 --> 00:57:01,180
We've got to show we can
color it in d plus 1 colors.

948
00:57:01,180 --> 00:57:03,120
Well the basic
algorithm, let's say.

949
00:57:03,120 --> 00:57:05,680
First thing we do is
we order the nodes

950
00:57:05,680 --> 00:57:06,954
in an arbitrary order.

951
00:57:10,363 --> 00:57:13,700
And we're going to show
whatever order you pick is OK.

952
00:57:22,920 --> 00:57:28,185
All right so what are the nodes?

953
00:57:38,740 --> 00:57:41,200
Anyway at all.

954
00:57:41,200 --> 00:57:45,690
Now how am I going to use
the induction hypothesis?

955
00:57:45,690 --> 00:57:49,900
I know, I can assume,
the for any N node graph

956
00:57:49,900 --> 00:57:54,550
I can color it in the
max degree plus 1 colors.

957
00:57:54,550 --> 00:57:56,940
How am I going to use that
to help me color G here,

958
00:57:56,940 --> 00:57:58,260
the n plus 1 node graph?

959
00:57:58,260 --> 00:58:00,690
Any thoughts?

960
00:58:00,690 --> 00:58:01,190
Yeah?

961
00:58:01,190 --> 00:58:03,860
AUDIENCE: [INAUDIBLE]

962
00:58:03,860 --> 00:58:06,920
PROFESSOR: Yeah, let's
create an n node graph

963
00:58:06,920 --> 00:58:09,410
by looking at these
nodes, and taking

964
00:58:09,410 --> 00:58:12,310
this one out of the time being.

965
00:58:12,310 --> 00:58:15,910
Remove the last V n plus
1 node in the order.

966
00:58:15,910 --> 00:58:19,400
That leaves an n node graph.

967
00:58:19,400 --> 00:58:22,310
So let's write that down.

968
00:58:22,310 --> 00:58:33,120
We remove the n plus 1 from G.
And that creates a new graph,

969
00:58:33,120 --> 00:58:38,530
call it G prime with vertices,
V prime and edges, E prime.

970
00:58:38,530 --> 00:58:40,925
So we create a new graph
by removing that node.

971
00:58:43,640 --> 00:58:46,240
And we remove all the
edges tied to that node.

972
00:58:46,240 --> 00:58:50,360
So for example over
here, the last node

973
00:58:50,360 --> 00:58:56,150
was 6042, so we take out
6042, and all these edges.

974
00:58:56,150 --> 00:58:59,116
And this is a graph
that we're left with.

975
00:59:02,370 --> 00:59:06,190
That graph has n nodes.

976
00:59:06,190 --> 00:59:08,637
What's the maximum
degree in G prime?

977
00:59:11,910 --> 00:59:18,054
When I pull out a node, can
the degree of any node go up?

978
00:59:18,054 --> 00:59:21,320
No, I'm just taking stuff out.

979
00:59:21,320 --> 00:59:27,930
So I know that G prime has
maximum degree, at most, d.

980
00:59:27,930 --> 00:59:30,370
The degree didn't
go up of any node.

981
00:59:35,080 --> 00:59:38,430
Might have gone down,
but it didn't go up.

982
00:59:38,430 --> 00:59:44,045
So G prime has max degree, at
most, d, and it has n nodes.

983
00:59:47,440 --> 00:59:50,440
So we can use the
induction hypothesis P n.

984
00:59:53,360 --> 01:00:01,320
It says that the basic
algorithm uses d plus 1,

985
01:00:01,320 --> 01:00:09,758
at most, d plus 1 colors
for nodes V1 to V n.

986
01:00:13,104 --> 01:00:14,260
Any questions about that?

987
01:00:17,140 --> 01:00:22,140
So if this were the n plus first
node, last node in the ordering

988
01:00:22,140 --> 01:00:24,090
take it out.

989
01:00:24,090 --> 01:00:26,490
The basic algorithm now,
take the same order here,

990
01:00:26,490 --> 01:00:33,850
V1, V2, V3, V4, basic, we'll
color that in d plus 1 colors.

991
01:00:33,850 --> 01:00:36,950
And all I have left is
to give this guy color,

992
01:00:36,950 --> 01:00:41,360
and I'll have color G. Question?

993
01:00:41,360 --> 01:00:43,640
No.

994
01:00:43,640 --> 01:00:45,600
All right.

995
01:00:45,600 --> 01:00:48,820
So by induction I've colored
these guys, V1 to V2,

996
01:00:48,820 --> 01:00:51,760
and d plus 1 colors, all
that I have left to do

997
01:00:51,760 --> 01:00:56,000
is color V n plus 1.

998
01:00:56,000 --> 01:00:59,585
And hopefully we're not
going to use color d plus 2,

999
01:00:59,585 --> 01:01:02,450
because then we sort
of-- it wouldn't work.

1000
01:01:02,450 --> 01:01:04,953
We got to use one of
the first d plus 1.

1001
01:01:12,330 --> 01:01:14,370
All right, so let's
look at V n plus 1.

1002
01:01:17,110 --> 01:01:25,780
And let's call its
neighbors in G, U1, U2, Ud.

1003
01:01:25,780 --> 01:01:31,630
It has, at most d neighbors,
because every node in G has,

1004
01:01:31,630 --> 01:01:34,320
at most, degree d.

1005
01:01:34,320 --> 01:01:35,980
A neighbor's a node
you're adjacent to.

1006
01:01:38,840 --> 01:01:45,950
All right so, V n plus 1 has at
most d neighbors, is adjacent

1007
01:01:45,950 --> 01:01:48,061
to, at most, d other nodes.

1008
01:01:51,220 --> 01:01:54,470
Now what does that
mean about the color I

1009
01:01:54,470 --> 01:01:56,990
can use on V n plus 1?

1010
01:01:59,540 --> 01:02:04,605
What do I know about what
color I can use for that?

1011
01:02:04,605 --> 01:02:05,105
Yeah?

1012
01:02:05,105 --> 01:02:10,070
AUDIENCE: It can't be any of
the colors of U1, U2, and so on.

1013
01:02:10,070 --> 01:02:12,850
PROFESSOR: It can't be any
one of these colors that

1014
01:02:12,850 --> 01:02:14,860
were assigned here.

1015
01:02:14,860 --> 01:02:16,110
That's true.

1016
01:02:16,110 --> 01:02:19,440
So how many colors
got ruled out?

1017
01:02:19,440 --> 01:02:23,300
At most d, and how
many am I working with?

1018
01:02:23,300 --> 01:02:25,560
d Plus 1.

1019
01:02:25,560 --> 01:02:29,580
So I got one left
that I can use safely.

1020
01:02:29,580 --> 01:02:30,920
OK.

1021
01:02:30,920 --> 01:02:35,660
So this means there
exists at least one color

1022
01:02:35,660 --> 01:02:37,890
in my set of d plus 1 colors.

1023
01:02:42,560 --> 01:02:47,575
It's not used by any neighbor.

1024
01:02:53,660 --> 01:02:56,390
And we're going to give
V n plus 1 that color.

1025
01:03:07,150 --> 01:03:07,650
All right.

1026
01:03:07,650 --> 01:03:12,350
So now I've colored every node
in G, the n plus 1 node graph,

1027
01:03:12,350 --> 01:03:15,315
safely using a total
of d plus 1 colors.

1028
01:03:18,692 --> 01:03:25,050
So that means the
basic algorithm uses,

1029
01:03:25,050 --> 01:03:33,340
at most, d plus 1 colors,
on G. That means P n plus 1

1030
01:03:33,340 --> 01:03:37,815
is true-- whoops-- and
the induction is complete.

1031
01:03:43,050 --> 01:03:44,853
Any questions?

1032
01:03:44,853 --> 01:03:45,353
Yeah.

1033
01:03:45,353 --> 01:03:49,281
AUDIENCE: Could you also
start from the other way,

1034
01:03:49,281 --> 01:03:55,664
and start 1, go to 2 nodes,
3 nodes at each step keeping

1035
01:03:55,664 --> 01:03:59,101
all nodes at all other nodes.

1036
01:03:59,101 --> 01:04:02,080
[INAUDIBLE]

1037
01:04:02,080 --> 01:04:05,308
PROFESSOR: What do you mean by
keeping all nodes connected?

1038
01:04:05,308 --> 01:04:16,264
AUDIENCE: [INAUDIBLE] each
node has an edge connecting

1039
01:04:16,264 --> 01:04:17,804
to each other one.

1040
01:04:17,804 --> 01:04:19,720
PROFESSOR: OK so, then
I get a specific graph.

1041
01:04:19,720 --> 01:04:24,240
I start with this, I add a
node and make it adjacent.

1042
01:04:24,240 --> 01:04:25,680
I add a node and
make it adjacent.

1043
01:04:25,680 --> 01:04:26,554
AUDIENCE: [INAUDIBLE]

1044
01:04:32,410 --> 01:04:33,904
PROFESSOR: Yeah.

1045
01:04:33,904 --> 01:04:36,620
So you've constructed
a particular graph.

1046
01:04:36,620 --> 01:04:40,260
This is actually called, for
the n nodes, it's called Kn,

1047
01:04:40,260 --> 01:04:45,010
is the n node complete
graph, also called a clique,

1048
01:04:45,010 --> 01:04:49,100
like a clique of friends, where
everybody likes everybody,

1049
01:04:49,100 --> 01:04:52,350
in a clique.

1050
01:04:52,350 --> 01:04:55,660
And in fact for n here,
for those n nodes,

1051
01:04:55,660 --> 01:04:58,940
what's the max degree?

1052
01:04:58,940 --> 01:05:01,440
Max degree is n minus 1.

1053
01:05:01,440 --> 01:05:03,420
What's the chromatic
number of this graph?

1054
01:05:05,986 --> 01:05:07,485
What's the minimum
number of colors?

1055
01:05:07,485 --> 01:05:09,156
[INTERPOSING VOICES]

1056
01:05:09,156 --> 01:05:11,030
PROFESSOR: And they all
have to be different,

1057
01:05:11,030 --> 01:05:12,890
which is d plus 1.

1058
01:05:12,890 --> 01:05:16,420
So you have built
a special graph

1059
01:05:16,420 --> 01:05:19,560
for which the optimum of
number colors is d plus 1.

1060
01:05:19,560 --> 01:05:23,630
But that is not a proof that
this is true for all graphs.

1061
01:05:23,630 --> 01:05:26,384
Because you've looked at
a particular graph here.

1062
01:05:26,384 --> 01:05:27,259
AUDIENCE: [INAUDIBLE]

1063
01:05:30,222 --> 01:05:31,180
PROFESSOR: What's that?

1064
01:05:31,180 --> 01:05:34,760
AUDIENCE: [INAUDIBLE] It means
that you can still use your

1065
01:05:34,760 --> 01:05:36,138
less than or equal to sign.

1066
01:05:36,138 --> 01:05:38,490
PROFESSOR: I see,
so you'd add a node,

1067
01:05:38,490 --> 01:05:41,182
and it's only connected
to a few of them.

1068
01:05:41,182 --> 01:05:43,015
AUDIENCE: No, it's
connected to all of them,

1069
01:05:43,015 --> 01:05:45,930
but it still implies
that you need less than

1070
01:05:45,930 --> 01:05:47,050
or equal to the colors.

1071
01:05:47,050 --> 01:05:49,083
It turns out it
happens to be equal to.

1072
01:05:49,083 --> 01:05:50,832
PROFESSOR: Yes, in
this case that's right.

1073
01:05:50,832 --> 01:05:52,510
So you've made an
argument for this case

1074
01:05:52,510 --> 01:05:54,060
where it actually
is equal, but that

1075
01:05:54,060 --> 01:05:55,200
only worked for this graph.

1076
01:05:55,200 --> 01:05:59,310
AUDIENCE: [INAUDIBLE]
worse case.

1077
01:05:59,310 --> 01:06:02,666
PROFESSOR: It is the worst
case, so it meets the bound.

1078
01:06:02,666 --> 01:06:04,290
It shows you cannot
improve this bound.

1079
01:06:04,290 --> 01:06:06,052
Yeah, is there a
question up there?

1080
01:06:06,052 --> 01:06:09,290
AUDIENCE: All I was going
to say is that you've

1081
01:06:09,290 --> 01:06:10,620
proved it's the worst case.

1082
01:06:10,620 --> 01:06:12,369
PROFESSOR: Right, so
what you've done here

1083
01:06:12,369 --> 01:06:15,587
is you've shown that I could not
make that theorem any stronger.

1084
01:06:15,587 --> 01:06:18,531
I could not replace
it with d here.

1085
01:06:18,531 --> 01:06:19,030
All right.

1086
01:06:19,030 --> 01:06:20,571
Because you've given
an example where

1087
01:06:20,571 --> 01:06:23,440
I can't get d colors, where
the maximum degree is d.

1088
01:06:23,440 --> 01:06:25,880
But that doesn't-- To get
a proof for a theorem,

1089
01:06:25,880 --> 01:06:27,110
I got to go through all this.

1090
01:06:27,110 --> 01:06:30,480
That wouldn't give me
a proof of the theorem.

1091
01:06:30,480 --> 01:06:32,130
They're not equivalent.

1092
01:06:32,130 --> 01:06:35,660
One's an upper bound, one's
an existence of a lower bound.

1093
01:06:35,660 --> 01:06:40,940
This shows that for any graph,
you need at most d plus 1.

1094
01:06:40,940 --> 01:06:43,990
So any graph, at most.

1095
01:06:43,990 --> 01:06:50,430
That shows there is a graph
that you need at least.

1096
01:06:50,430 --> 01:06:52,640
And they are not equivalent.

1097
01:06:52,640 --> 01:06:53,230
All right.

1098
01:06:53,230 --> 01:06:56,610
One is for all, and upper bound.

1099
01:06:56,610 --> 01:07:00,830
The other is there
exists a lower bound.

1100
01:07:00,830 --> 01:07:05,590
So different in two
ways that are important.

1101
01:07:05,590 --> 01:07:08,980
This kind of proof is very
typical for what you'll

1102
01:07:08,980 --> 01:07:11,390
see with induction in graphs.

1103
01:07:11,390 --> 01:07:13,340
And you'll get a lot
of practice with it.

1104
01:07:13,340 --> 01:07:17,380
Are there any other
questions on this proof?

1105
01:07:20,200 --> 01:07:21,870
OK.

1106
01:07:21,870 --> 01:07:24,800
All right, see we've seen
now, by that example,

1107
01:07:24,800 --> 01:07:27,330
we can't improve the theorem.

1108
01:07:27,330 --> 01:07:30,200
In some cases,
though, the theorem

1109
01:07:30,200 --> 01:07:32,020
is way off, for some graphs.

1110
01:07:32,020 --> 01:07:34,530
Can anybody think
of a graph where

1111
01:07:34,530 --> 01:07:37,850
the bound we get from the
theorem, of d plus 1 colors,

1112
01:07:37,850 --> 01:07:40,680
is way off from the actual
chromatic number you need,

1113
01:07:40,680 --> 01:07:41,962
the number of colors you need?

1114
01:07:41,962 --> 01:07:42,462
Yeah?

1115
01:07:42,462 --> 01:07:44,000
AUDIENCE: [INAUDIBLE]

1116
01:07:44,000 --> 01:07:44,916
PROFESSOR: What is it?

1117
01:07:44,916 --> 01:07:51,282
AUDIENCE: A graph [INAUDIBLE]
two sets of [INAUDIBLE]

1118
01:07:55,130 --> 01:07:56,470
PROFESSOR: Good, OK.

1119
01:07:56,470 --> 01:07:58,220
Yes, so what if
we did this graph.

1120
01:07:58,220 --> 01:08:00,042
Let me draw it out.

1121
01:08:03,310 --> 01:08:09,600
So you've got a bunch of nodes
here, bunch of nodes here.

1122
01:08:09,600 --> 01:08:12,300
And every node here is
connected to every node

1123
01:08:12,300 --> 01:08:15,600
over the other side.

1124
01:08:15,600 --> 01:08:18,050
And if this is an n no
graph, and I've got n over 2

1125
01:08:18,050 --> 01:08:21,100
on each side, what's
my degree here?

1126
01:08:25,922 --> 01:08:27,380
What's my max degree
of this graph?

1127
01:08:27,380 --> 01:08:28,560
AUDIENCE: N over 2.

1128
01:08:28,560 --> 01:08:30,319
PROFESSOR: N over 2.

1129
01:08:30,319 --> 01:08:32,680
So d is n over 2.

1130
01:08:32,680 --> 01:08:34,060
What's the chromatic number?

1131
01:08:34,060 --> 01:08:36,520
How many colors do
I need for this?

1132
01:08:36,520 --> 01:08:38,250
Two.

1133
01:08:38,250 --> 01:08:41,380
All right, so d plus
1 is way off of two.

1134
01:08:41,380 --> 01:08:43,466
There is a even worse example.

1135
01:08:43,466 --> 01:08:43,966
Yeah?

1136
01:08:43,966 --> 01:08:48,130
AUDIENCE: That graph where you
have one node center that's

1137
01:08:48,130 --> 01:08:51,126
connected to a bunch of nodes
regularly distributed about.

1138
01:08:51,126 --> 01:08:52,564
PROFESSOR: Yeah, the star graph.

1139
01:08:52,564 --> 01:08:56,300
All right, so I got one of
the center, I got n minus 1

1140
01:08:56,300 --> 01:08:58,330
outside.

1141
01:08:58,330 --> 01:09:03,050
So here the maximum
degree is n minus 1,

1142
01:09:03,050 --> 01:09:05,460
just like a complete graph.

1143
01:09:05,460 --> 01:09:08,649
But how many colors do I need?

1144
01:09:08,649 --> 01:09:10,069
Two.

1145
01:09:10,069 --> 01:09:11,500
So it's even worse here.

1146
01:09:14,319 --> 01:09:18,920
All right now what about
the basic algorithm?

1147
01:09:18,920 --> 01:09:21,599
How well does the basic
algorithm do on this graph?

1148
01:09:24,649 --> 01:09:27,790
Or to the vertices some way?

1149
01:09:27,790 --> 01:09:29,630
Color on one [INAUDIBLE]
lowest color.

1150
01:09:29,630 --> 01:09:31,620
How many colors is
it going to use?

1151
01:09:31,620 --> 01:09:33,840
AUDIENCE: Two.

1152
01:09:33,840 --> 01:09:35,240
PROFESSOR: Two.

1153
01:09:35,240 --> 01:09:38,109
It doesn't matter the vertices.

1154
01:09:38,109 --> 01:09:46,619
V1, V2, V3, V4, because
I'll color this one 1.

1155
01:09:46,619 --> 01:09:49,668
What am I going
to call that one?

1156
01:09:49,668 --> 01:09:51,950
1.

1157
01:09:51,950 --> 01:09:54,770
Then I get to the center,
what am I going to color it?

1158
01:09:54,770 --> 01:09:56,360
2.

1159
01:09:56,360 --> 01:10:00,130
And now all the arms,
what do they get colored?

1160
01:10:00,130 --> 01:10:01,240
They all get 1.

1161
01:10:01,240 --> 01:10:04,275
Whatever order you pick,
you get two colors.

1162
01:10:07,270 --> 01:10:08,892
All right so now
there's a difference

1163
01:10:08,892 --> 01:10:11,350
between the theorem just gives
you an upper bound, it says,

1164
01:10:11,350 --> 01:10:13,150
at most, d plus 1 colors.

1165
01:10:13,150 --> 01:10:15,700
But in fact the algorithm
can do a lot better

1166
01:10:15,700 --> 01:10:19,050
than that, as on this example.

1167
01:10:19,050 --> 01:10:22,420
So the algorithm
might be a lot better.

1168
01:10:22,420 --> 01:10:24,770
Everybody see that
what we're doing here?

1169
01:10:24,770 --> 01:10:26,520
How the algorithm is
better than the bound

1170
01:10:26,520 --> 01:10:28,644
we proved by the theorem,
even though the bound was

1171
01:10:28,644 --> 01:10:31,180
pretty good for some graphs.

1172
01:10:31,180 --> 01:10:33,620
Now it turns out--
I mean we're not

1173
01:10:33,620 --> 01:10:36,110
going to win a million
dollars for this algorithm.

1174
01:10:36,110 --> 01:10:39,880
And in fact, this algorithm
is sometimes very bad.

1175
01:10:39,880 --> 01:10:44,370
And a really bad example
it's very close to this.

1176
01:10:44,370 --> 01:10:48,470
In fact actually this one, let's
look at how well does basic

1177
01:10:48,470 --> 01:10:51,310
do one this one here.

1178
01:10:54,240 --> 01:10:56,160
Make some ordering.

1179
01:10:56,160 --> 01:10:59,950
V1, V2, V3.

1180
01:11:02,715 --> 01:11:05,520
What's the basic algorithm
going to do on this complete--

1181
01:11:05,520 --> 01:11:08,780
it's called a complete bipartite
graph, is what's this called.

1182
01:11:08,780 --> 01:11:10,300
I'll define bipartite
in a minute--

1183
01:11:10,300 --> 01:11:12,001
but what's the basic
algorithm do here?

1184
01:11:15,690 --> 01:11:20,860
Any idea-- does it take n over
2 colors, or does it take 2?

1185
01:11:20,860 --> 01:11:23,170
Any ideas?

1186
01:11:23,170 --> 01:11:24,370
2.

1187
01:11:24,370 --> 01:11:29,650
So take a vertex, and the first
one, say V1s here, get C1.

1188
01:11:29,650 --> 01:11:32,790
As long as I keep picking
vertices over on this side,

1189
01:11:32,790 --> 01:11:33,790
they're going to get C1.

1190
01:11:36,430 --> 01:11:39,560
As soon as I get to
a vertex over here,

1191
01:11:39,560 --> 01:11:41,363
what color does it have to get?

1192
01:11:41,363 --> 01:11:42,130
AUDIENCE: C2.

1193
01:11:42,130 --> 01:11:43,620
PROFESSOR: C2
because it's touching

1194
01:11:43,620 --> 01:11:46,296
the very first one we had here.

1195
01:11:46,296 --> 01:11:47,670
So when I get
vertices over here,

1196
01:11:47,670 --> 01:11:48,795
they're all going to be C2.

1197
01:11:48,795 --> 01:11:52,270
When I go back over here,
they're going to be back to C1.

1198
01:11:52,270 --> 01:11:55,910
So actually basic does good
here too, gives you two colors.

1199
01:11:55,910 --> 01:11:56,410
Yeah?

1200
01:11:56,410 --> 01:11:57,408
AUDIENCE: [INAUDIBLE]

1201
01:12:06,870 --> 01:12:08,620
PROFESSOR: Ah, those
two aren't connected.

1202
01:12:08,620 --> 01:12:11,480
But this case, if I've
got a vertex over here

1203
01:12:11,480 --> 01:12:14,540
it is, by definition, connected
to the vertex over here.

1204
01:12:14,540 --> 01:12:17,820
Because every
possible edge is here.

1205
01:12:17,820 --> 01:12:18,870
But that's a great idea.

1206
01:12:18,870 --> 01:12:22,447
What if they weren't
all connected,

1207
01:12:22,447 --> 01:12:23,655
that's actually a great idea.

1208
01:12:25,590 --> 01:12:29,270
In fact, the nasty example
for the basic algorithm

1209
01:12:29,270 --> 01:12:30,270
is very much like that.

1210
01:12:30,270 --> 01:12:31,446
Let's draw it.

1211
01:12:35,748 --> 01:12:38,200
Because so far,
the basic algorithm

1212
01:12:38,200 --> 01:12:42,080
is pretty much done
perfectly on all the graphs

1213
01:12:42,080 --> 01:12:48,870
we looked at even when
the theorem wasn't tight.

1214
01:12:48,870 --> 01:12:50,445
So here is a nasty graph.

1215
01:12:56,140 --> 01:12:58,930
And it is very close to the
graph we just look like,

1216
01:12:58,930 --> 01:13:00,910
where all the edges are there.

1217
01:13:00,910 --> 01:13:03,080
In this case, all
the edges are there,

1218
01:13:03,080 --> 01:13:07,040
except for the one
straight across.

1219
01:13:07,040 --> 01:13:10,980
So if this is-- the
edge denotes likes,

1220
01:13:10,980 --> 01:13:14,410
this is a world where you like
everybody but your spouse.

1221
01:13:16,950 --> 01:13:19,390
All right, so you have
an edge to every one,

1222
01:13:19,390 --> 01:13:21,240
except the one directly
across from you.

1223
01:13:25,190 --> 01:13:29,370
No edge there, and so forth.

1224
01:13:29,370 --> 01:13:32,460
So it has almost every edge,
but it's missing these edges.

1225
01:13:37,620 --> 01:13:42,440
Now the basic algorithm
might do well here.

1226
01:13:42,440 --> 01:13:46,120
What would be a good
ordering for this graph

1227
01:13:46,120 --> 01:13:48,010
to label these V1 through Vn?

1228
01:13:48,010 --> 01:13:48,510
Yeah?

1229
01:13:48,510 --> 01:13:51,630
AUDIENCE: Go through
everything on the left side,

1230
01:13:51,630 --> 01:13:54,818
and then the right side.

1231
01:13:54,818 --> 01:13:56,312
PROFESSOR: Yeah, that's right.

1232
01:14:03,800 --> 01:14:12,480
Because then color 1, color
1, color 1, all the way down.

1233
01:14:12,480 --> 01:14:17,460
One color for the left,
what does this one get?

1234
01:14:17,460 --> 01:14:19,540
Color 2, because it's
hooked up against.

1235
01:14:19,540 --> 01:14:25,670
And these all get color 2,
so I've used two colors.

1236
01:14:25,670 --> 01:14:26,260
Really good.

1237
01:14:26,260 --> 01:14:28,630
Basic algorithm's looking great.

1238
01:14:28,630 --> 01:14:31,470
Now here's a harder question.

1239
01:14:31,470 --> 01:14:34,800
Can you figure
out a bad ordering

1240
01:14:34,800 --> 01:14:37,995
for this graph, where I use
a lot more than two colors.

1241
01:14:37,995 --> 01:14:38,870
AUDIENCE: [INAUDIBLE]

1242
01:14:42,124 --> 01:14:43,040
PROFESSOR: What is it?

1243
01:14:43,040 --> 01:14:47,286
AUDIENCE: It starts at the
top of the cross, and then

1244
01:14:47,286 --> 01:14:48,685
the next level then across.

1245
01:14:48,685 --> 01:14:49,560
PROFESSOR: Very good.

1246
01:14:49,560 --> 01:14:50,740
V1, V2.

1247
01:14:50,740 --> 01:14:53,210
Just as natural, really,
if think about it,

1248
01:14:53,210 --> 01:14:54,490
to order it this way.

1249
01:15:02,670 --> 01:15:03,860
All right.

1250
01:15:03,860 --> 01:15:06,650
What color does V1 get?

1251
01:15:06,650 --> 01:15:07,410
C1.

1252
01:15:07,410 --> 01:15:08,890
What color does V2 get?

1253
01:15:08,890 --> 01:15:09,910
AUDIENCE: C1.

1254
01:15:09,910 --> 01:15:12,250
PROFESSOR: C1 because
it's not hooked up here.

1255
01:15:12,250 --> 01:15:13,735
What color does V3 get?

1256
01:15:13,735 --> 01:15:14,387
AUDIENCE: C2.

1257
01:15:14,387 --> 01:15:14,970
PROFESSOR: C2.

1258
01:15:14,970 --> 01:15:16,480
What about V4?

1259
01:15:16,480 --> 01:15:17,887
AUDIENCE: C2.

1260
01:15:17,887 --> 01:15:18,470
PROFESSOR: C2.

1261
01:15:18,470 --> 01:15:19,670
It's not hooked up.

1262
01:15:19,670 --> 01:15:21,560
It can't get one,
because that's up here.

1263
01:15:21,560 --> 01:15:25,310
And it's not the two, so it
gets two What color does V5 get?

1264
01:15:25,310 --> 01:15:26,600
AUDIENCE: C3.

1265
01:15:26,600 --> 01:15:27,490
PROFESSOR: C3.

1266
01:15:27,490 --> 01:15:29,450
Because it's hooked
up to one to two.

1267
01:15:29,450 --> 01:15:30,806
V6 ?

1268
01:15:30,806 --> 01:15:31,610
AUDIENCE: C3.

1269
01:15:31,610 --> 01:15:34,900
PROFESSOR: C3, it's hooked up
to one and two, but not three.

1270
01:15:34,900 --> 01:15:38,560
And you can see
what's happening here.

1271
01:15:38,560 --> 01:15:40,230
All the way down
here he's hooked up

1272
01:15:40,230 --> 01:15:43,165
to all the n over
2 minus 1 colors.

1273
01:15:43,165 --> 01:15:47,110
So he also takes C n over 2.

1274
01:15:47,110 --> 01:15:50,120
So if you pick that
ordering, not so good.

1275
01:15:50,120 --> 01:15:52,600
You use n over two colors.

1276
01:15:52,600 --> 01:15:55,510
So it really matters
the ordering.

1277
01:15:55,510 --> 01:15:58,030
Now I should say graphs
like-- actually any questions

1278
01:15:58,030 --> 01:15:59,700
about what we did here?

1279
01:15:59,700 --> 01:16:02,430
About this?

1280
01:16:02,430 --> 01:16:04,560
All right, now I should
say that graphs like this

1281
01:16:04,560 --> 01:16:09,470
have a special name, they're
called bipartite graphs.

1282
01:16:09,470 --> 01:16:11,249
And that's important
to remember.

1283
01:16:24,520 --> 01:16:43,780
All right, so a graph G
is said to be bipartite

1284
01:16:43,780 --> 01:16:53,316
if the vertices can be split
into two sets, or partitioned,

1285
01:16:53,316 --> 01:17:01,360
and we'll call them a
left set, and a right set,

1286
01:17:01,360 --> 01:17:17,330
so that all the edges connect
a node in the left set,

1287
01:17:17,330 --> 01:17:21,190
to a node in the right set.

1288
01:17:23,759 --> 01:17:25,300
So in fact, a lot
of today we've been

1289
01:17:25,300 --> 01:17:28,500
looking at bipartite graphs,
because the nodes are here.

1290
01:17:28,500 --> 01:17:32,660
Like the men, and the women, and
the edges only go from the left

1291
01:17:32,660 --> 01:17:33,900
to the right.

1292
01:17:33,900 --> 01:17:35,335
And that is called bipartite.

1293
01:17:35,335 --> 01:17:37,290
And it's called
bipartite because you

1294
01:17:37,290 --> 01:17:41,820
can do it with two
colors, or in two pieces.

1295
01:17:41,820 --> 01:17:43,860
So you don't win
a million dollars

1296
01:17:43,860 --> 01:17:45,630
for deciding whether
or not a graph can

1297
01:17:45,630 --> 01:17:47,600
be colored in two colors.

1298
01:17:47,600 --> 01:17:48,950
That's easy.

1299
01:17:48,950 --> 01:17:51,730
You'll even do it for
homework one of these times.

1300
01:17:51,730 --> 01:17:54,330
You do win the million dollars
for deciding if a graph can

1301
01:17:54,330 --> 01:17:56,700
be colored in three colors.

1302
01:17:56,700 --> 01:17:58,450
That's really hard to do.

1303
01:18:01,090 --> 01:18:04,800
Now coloring problems come up
in all sorts of applications.

1304
01:18:04,800 --> 01:18:08,950
You know with this company,
Akamai, that came out of MIT,

1305
01:18:08,950 --> 01:18:10,070
we've talked about.

1306
01:18:10,070 --> 01:18:14,490
We run a network
of 75,000 servers.

1307
01:18:14,490 --> 01:18:16,860
And they're used to distribute
content on the internet,

1308
01:18:16,860 --> 01:18:18,280
and so forth.

1309
01:18:18,280 --> 01:18:20,210
And we have to
deploy a new version

1310
01:18:20,210 --> 01:18:21,800
of our software
on those servers,

1311
01:18:21,800 --> 01:18:23,420
pretty much every week.

1312
01:18:23,420 --> 01:18:25,560
We're pushing new software out.

1313
01:18:25,560 --> 01:18:31,070
And you can't deploy on every
server at the same time,

1314
01:18:31,070 --> 01:18:32,950
because you've got
to take down a server

1315
01:18:32,950 --> 01:18:34,450
to deploy new software on it.

1316
01:18:34,450 --> 01:18:36,050
Got to take it
out of commission.

1317
01:18:36,050 --> 01:18:38,650
And so we can't just take
down all 75,000 servers,

1318
01:18:38,650 --> 01:18:40,960
because then all the
Facebook, and Netflix,

1319
01:18:40,960 --> 01:18:42,410
and all those sites would stop.

1320
01:18:42,410 --> 01:18:44,840
That would be bad.

1321
01:18:44,840 --> 01:18:48,060
And we can't do them one at a
time, because there's 75,000.

1322
01:18:48,060 --> 01:18:50,780
And it takes a few
hours for each one

1323
01:18:50,780 --> 01:18:54,070
to get the traffic off,
stop it, load new software,

1324
01:18:54,070 --> 01:18:54,980
and turn it back on.

1325
01:18:54,980 --> 01:18:58,240
And it would take us years
to do one software install,

1326
01:18:58,240 --> 01:19:00,610
which we got to do every week.

1327
01:19:00,610 --> 01:19:03,280
So we've got to figure out a
schedule for how many servers

1328
01:19:03,280 --> 01:19:05,530
you take down at a given
time, and which ones.

1329
01:19:05,530 --> 01:19:07,590
And it turns out
pairs of servers

1330
01:19:07,590 --> 01:19:09,639
have certain critical functions.

1331
01:19:09,639 --> 01:19:11,930
So there's certain pairs of
servers you can't take down

1332
01:19:11,930 --> 01:19:14,750
at the same time.

1333
01:19:14,750 --> 01:19:18,536
So we have a gigantic 75,000
node coloring problem,

1334
01:19:18,536 --> 01:19:20,035
where there's edges
between servers.

1335
01:19:20,035 --> 01:19:21,618
Nodes are servers,
and there's an edge

1336
01:19:21,618 --> 01:19:25,634
between if you can't install
new software at the same time.

1337
01:19:25,634 --> 01:19:27,050
And so when it
turns out, when you

1338
01:19:27,050 --> 01:19:30,410
run one of these graph
coloring algorithms on it,

1339
01:19:30,410 --> 01:19:32,120
you could do it
with eight colors.

1340
01:19:32,120 --> 01:19:34,160
It just turns out that way.

1341
01:19:34,160 --> 01:19:37,070
So that means there's
eight waves of install

1342
01:19:37,070 --> 01:19:38,920
that go on to the network.

1343
01:19:38,920 --> 01:19:41,450
And now eight times
a few hours each

1344
01:19:41,450 --> 01:19:44,672
means that we can do it in a
day, and you can manage it.

1345
01:19:44,672 --> 01:19:47,780
You know on a much
smaller scale,

1346
01:19:47,780 --> 01:19:50,310
the same problem exists
for register allocation,

1347
01:19:50,310 --> 01:19:52,120
for variables.

1348
01:19:52,120 --> 01:19:55,486
Here you've got to assign
every variable to register.

1349
01:19:55,486 --> 01:19:56,860
But you can't have
variables that

1350
01:19:56,860 --> 01:19:59,360
are active at the
same time associated

1351
01:19:59,360 --> 01:20:00,900
with the same register.

1352
01:20:00,900 --> 01:20:03,850
And you want to minimize the
number of registers you need.

1353
01:20:03,850 --> 01:20:07,110
So again, you have the
graph coloring problem.

1354
01:20:07,110 --> 01:20:10,220
The number of colors is the
number of registers you need.

1355
01:20:10,220 --> 01:20:12,920
And two variables can't get
the same color if their active

1356
01:20:12,920 --> 01:20:15,810
at the same time, so you
put an edge between them.

1357
01:20:15,810 --> 01:20:17,720
The most famous example
of graph coloring

1358
01:20:17,720 --> 01:20:21,670
is the map coloring problem,
with the four coloring theorem.

1359
01:20:21,670 --> 01:20:25,800
And so here, every
country is a node.

1360
01:20:25,800 --> 01:20:27,727
Adjacent countries have
an edge between them,

1361
01:20:27,727 --> 01:20:29,810
because you don't want to
color adjacent countries

1362
01:20:29,810 --> 01:20:31,480
the same color,
or you can't tell

1363
01:20:31,480 --> 01:20:34,730
they're different countries.

1364
01:20:34,730 --> 01:20:36,990
Now the last example
we can talk about

1365
01:20:36,990 --> 01:20:40,000
is an important problem
in communication theory,

1366
01:20:40,000 --> 01:20:43,630
communication networks, where
again coloring comes up.

1367
01:20:43,630 --> 01:20:48,100
Now here you need to assign
frequencies to radio stations,

1368
01:20:48,100 --> 01:20:49,035
or the cell towers.

1369
01:20:49,035 --> 01:20:54,510
It comes up in mobile networks,
or just in with radio stations.

1370
01:20:54,510 --> 01:20:59,370
And if two towers have
an overlapping area,

1371
01:20:59,370 --> 01:21:01,290
they can't be given
the same frequency,

1372
01:21:01,290 --> 01:21:05,540
so you get collisions
between the towers.

1373
01:21:05,540 --> 01:21:07,110
And frequencies
are very expensive.

1374
01:21:07,110 --> 01:21:08,901
Companies pay the
government a lot of money

1375
01:21:08,901 --> 01:21:11,020
to get certain spectrum.

1376
01:21:11,020 --> 01:21:13,350
So suppose you had this problem.

1377
01:21:13,350 --> 01:21:17,205
Here's tower A, this is A's
range, where it reaches.

1378
01:21:17,205 --> 01:21:20,750
Here's tower B, so
it overlaps some

1379
01:21:20,750 --> 01:21:36,660
with A. Here's tower C. Here's
tower E. And here's tower D.

1380
01:21:36,660 --> 01:21:38,280
All right now the
question would be,

1381
01:21:38,280 --> 01:21:40,869
how many radio
frequencies do you need?

1382
01:21:40,869 --> 01:21:42,910
What's the minimum number
of frequencies you need

1383
01:21:42,910 --> 01:21:45,916
to enable all the towers here?

1384
01:21:45,916 --> 01:21:47,480
We could make that be a graph.

1385
01:21:47,480 --> 01:21:49,760
There's a node for each tower.

1386
01:21:49,760 --> 01:21:54,400
And an edge between
towers, if they overlap.

1387
01:21:54,400 --> 01:21:58,650
C doesn't overlap
with B, E does.

1388
01:21:58,650 --> 01:22:00,970
E overlaps here.

1389
01:22:00,970 --> 01:22:05,060
And then D overlaps here.

1390
01:22:05,060 --> 01:22:09,197
So how many frequencies do
you need for this graph?

1391
01:22:09,197 --> 01:22:10,360
AUDIENCE: Four.

1392
01:22:10,360 --> 01:22:14,580
PROFESSOR: Four would
work, three is better.

1393
01:22:14,580 --> 01:22:16,512
Can you do two?

1394
01:22:19,900 --> 01:22:22,380
No you can't do two,
because you got here.

1395
01:22:22,380 --> 01:22:23,820
But you could do three.

1396
01:22:23,820 --> 01:22:29,830
You could do one,
two, three, two, one.

1397
01:22:29,830 --> 01:22:30,841
This problem comes up--

1398
01:22:30,841 --> 01:22:31,715
AUDIENCE: [INAUDIBLE]

1399
01:22:31,715 --> 01:22:33,010
PROFESSOR: Did I screw up?

1400
01:22:33,010 --> 01:22:34,660
Ooh, no I can't do that.

1401
01:22:34,660 --> 01:22:36,630
One, two, yeah much better.

1402
01:22:36,630 --> 01:22:39,300
All right, this problem
comes up all over the place.

1403
01:22:39,300 --> 01:22:41,640
I'm certain you'll see it
sometime in your career,

1404
01:22:41,640 --> 01:22:44,015
you'll have some problem, or
you're scheduling something,

1405
01:22:44,015 --> 01:22:45,960
and it's really a graph
problem in disguise.

1406
01:22:45,960 --> 01:22:48,250
OK that's it for today.