Placeholder Image

字幕列表 影片播放

  • The following content is provided under a Creative

  • Commons license.

  • Your support will help MIT OpenCourseWare

  • continue to offer high quality educational resources for free.

  • To make a donation or view additional materials

  • from hundreds of MIT courses, visit MIT OpenCourseWare

  • at ocw.mit.edu.

  • PROFESSOR: OK let's get started.

  • I once taught with a professor who

  • was lamenting the fact that as the term progresses attendance

  • in lecture tends to drop off.

  • And gets pretty dramatic by the end of the term

  • when you're lecturing, and nobody's there.

  • And I asked him what he did about it.

  • And he thought about it and he said,

  • there's only two things that can get students to come

  • to lecture, candy and sex.

  • Now we've already tried candy, so today we're

  • going to talk about sex.

  • In fact we're going to use graph theory

  • to address a decades old debate concerning

  • the relative promiscuity of men versus women.

  • Now graphs are incredibly useful structures in computer science,

  • and we're going to be studying them for the next five or six

  • lectures.

  • They come up in all sorts of applications, scheduling,

  • optimization, communications, the design

  • and analysis of algorithms.

  • In fact next week, you're going to see how to Stanford

  • graduate students became gazillionaires

  • because they use graph theoretic in a clever way.

  • But let's talk about sex.

  • The issue that we're going to address today

  • is one of the most talked about, and most well studied,

  • questions in all of human sociology.

  • On average, who has more opposite gender partners,

  • men or women?

  • Now opposite gender is going to be important.

  • And by this I mean, one boy, and one girl.

  • All right, I'm not making a political statement.

  • It's just that the math is a lot easier that way, as you'll see.

  • Now I'd like to start by taking a pole here

  • to see what you think about that.

  • So raise your and if you think men, on average,

  • have more opposite gender partners than women do.

  • Only a few.

  • AUDIENCE: In life or [INAUDIBLE]

  • PROFESSOR: Um, you can--

  • [LAUGHTER]

  • PROFESSOR: One on one.

  • OK, so let's say over the course of their lives, let's say,

  • or over the course of 2010, that men in America

  • have more opposite-gender partners than women in America,

  • say in 2010.

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

  • All right a bunch of you.

  • Raise your hand if you think women have

  • more opposite-gender partners?

  • This is unusual.

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

  • Raise your hand if you think it's equal.

  • All right, about the same.

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

  • that it's hopeless to really figure it out.

  • All right, nobody goes for that.

  • All right, good.

  • All right well now in the popular literature,

  • I think the feelings are different than expressed here.

  • Pretty much universally, in the literature,

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

  • than women.

  • And in fact, you could even think

  • about that, if you think about literature,

  • the leader of the harem is always a man.

  • And he's got lots of women.

  • In polygamist cultures, it's always

  • the man that has multiple wives, not the reverse.

  • Now not surprisingly, this issue has

  • been studied "scientifically," I'll

  • put in quotes, extensively, in one of the largest studies ever

  • done.

  • Researchers from University of Chicago

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

  • They brought them in, on many occasions,

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

  • And they wrote this 700 page book,

  • called The soul of Social Organization of Sexuality:

  • Sexual Practices in the US.

  • Actually walking around with this book

  • has proved to be a little embarrassing.

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

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

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

  • I'm teaching.

  • And I thought I'd gotten away with it,

  • and everything was fine.

  • And then later that day she texted all of our friends

  • about the new news that what do you know,

  • her dad teaches sex ed at MIT.

  • Anyway this study concludes that on average men

  • have 74% more opposite-gender partners than women.

  • There's one other central claims.

  • And this is in the US.

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

  • might be OK.

  • But not according to ABC News.

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

  • and concluded that the average disparity is much greater.

  • In particular, in this study, they

  • said that the average man has 20 partners--

  • I'm assuming over their lifetime--

  • and the average woman has six.

  • And this gives a disparity 233%.

  • So ABC News did a smaller survey says that it's 233% here,

  • much more than 74%.

  • Now ABC News claimed this is one of the most

  • scientific studies ever done.

  • And there was a 2.5% margin of error.

  • Now we'll actually talk about what that means mathematically

  • later in the term when we do probability, and do

  • study polling.

  • Now of course I should also mention

  • that ABC News is the one that said

  • Al Gore won the presidential election in 2000.

  • Now the study is called American Sex Survey,

  • a Peak Between the Sheets.

  • That doesn't sound so scientific.

  • And it was on TV, on Primetime Live in 2004.

  • The promo for this is really good.

  • It says, a groundbreaking ABC News Primetime Live survey

  • finds a range of eye popping sexual activities, fantasies,

  • and attitudes in this country, confirming

  • some conventional wisdom, exploding some myths,

  • and venturing where few scientific surveys have

  • gone before.

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

  • with that last statement.

  • OK now who do you think's right?

  • University of Chicago.

  • Who votes for 74% as being pretty close?

  • A few of you.

  • I've already slammed these guys.

  • Who votes for ABC News as being more accurate?

  • Yeah, nobody.

  • Who votes for no way to tell?

  • I got some votes there, all right.

  • So how do you tackle this problem?

  • In theory we could do our own 6.042 survey.

  • I don't know how much we'd really learn,

  • and for sure I'd get fired.

  • So I don't think we're going to do that.

  • But fortunately, this is the kind of question

  • that could be handled, and actually

  • answered, by graph theory, even though it might

  • be more interesting to interview thousands of people,

  • and find out what's going on.

  • That's not as efficient as using graphs.

  • So let me start by defining what a graph is.

  • Informally graph is just a bunch of dots and lines

  • connecting the dots, it's actually very simple.

  • So here's to graph.

  • These are the nodes, and they're connected

  • with these lines, called edges.

  • And often the nodes, and sometimes the edges,

  • are labeled.

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

  • So that's an example of a graph.

  • Now this being a math class, we got

  • to give a formal definition of a graph.

  • And we'll usually use the formal definition.

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

  • Where V is a set of elements called vertices or nodes.

  • And it has to be non-empty here in this class.

  • And we'll go back and forth between vertices and nodes.

  • Even the text we use both words interchangeably.

  • And E is a set of 2-item subsets and V,

  • and they're called edges.

  • So for example, over here in this picture,

  • V is the set of nodes is x1, x2, x3, up to x7, that's the nodes.

  • And E, the set of edges, is pairs,

  • unordered pairs of vertices.

  • So for example x1, x2 is an edge.

  • And it's the same as the set x2, x1,

  • doesn't matter the order here.

  • Later in a week or so, we'll talk about directed graphs

  • where the order matters.

  • x1, x3 is also an edge here, and so one.

  • Think we've got, let's see, 1, 2, 3, 4, 5, 6,

  • 7 edges in this graph.

  • And the last one would be x5, x7.

  • Edges are also sometimes written with this notation, x1 line x2,

  • is another notation.

  • And then later when you talk about directed edges,

  • we'll put a little arrowhead on one end of this.

  • Now the definition of a graph is really pretty simple.

  • Just think of it as dots and lines, if you want.

  • But there's often differences in how people define graphs.

  • For example, in this class we don't allow the empty graph,

  • i.e. the graph with no nodes.

  • So we're going to insist that every graph has to have

  • at least one node in it.

  • And that's just to make the theorems

  • we're going prove be true.

  • Otherwise there's some theorems that

  • are false for the special case of the empty graph.

  • But we don't require the graph to have any edges.

  • In fact, it's possible you have a graph with nodes,

  • but no edges.

  • For example, this graph.

  • Three-node graph.

  • So here G equals VE, V equals x1, x2, x3.

  • And E is just the empty set.

  • Now for a general graph, when you do have edges,

  • we say that two nodes, call them xi and xj,

  • are adjacent if they're connected by an edge, namely

  • if xi xj is an edge.

  • All right so for example, x5 is adjacent to x7,

  • but it's not adjacent to x4, there's no edge there.

  • Closely related is the definition of the incidence.

  • An edge E, which is xixj, is said

  • to be incident to its end points, xi and xj.

  • OK so, for example, if I labeled that edge as E, E

  • is the edge x1, x2, and this incident to x1, and incident

  • x2.

  • Then we can talk about the degree of a node.

  • The number of edges incident to a node

  • is called the degree of the node.

  • So for example, what's the degree of x5 over here?

  • 3, so in this case, the degree of x5 equals 3.

  • The degree of x7 is 1.

  • These guys all have degree 0, there's

  • no edges incident to them.

  • Now in this class, we're going to look at only simple graphs,

  • at lease for a while.

  • A graph is simple if it has no loops, or multiple edges.

  • Now a loop is an edge that only connects up one node, that's

  • a loop and we don't allow it.

  • A multiple edge is we've got two edges that are really the same,

  • they connect the same endpoints.

  • Also called a multi-edge.

  • And those we're not going to have in simple graphs.

  • We don't allow this.

  • We don't allow that.

  • Any questions so far about what a graph is?

  • So how are we going to use a graph

  • to model the problem of opposite-gender partners?

  • That's the question we're after.

  • So any thoughts about what the nodes of the graph

  • are going to represent?

  • What is it?

  • AUDIENCE: Males and females?

  • PROFESSOR: People.

  • Yeah, so we're going to have people.

  • In fact, there's two kinds of people here.

  • There's men, and women.

  • All right we got nodes here for the men.

  • And in fact in America, there's a lot of nodes here.

  • All right, and so this might be oh I don't know,

  • say that's Tom Cruise and Nicole Kidman.

  • Now what's the edge going to represent?

  • AUDIENCE: Partners.

  • PROFESSOR: Partners.

  • They were opposite-gender partners.

  • And there's actually more edges probably here.

  • We could have Penelope here, and Katie here.

  • And well probably lots more, I probably don't know them all.

  • And Ben's over here with Nicole.

  • And Nicole got Jude and Keith.

  • There's actually a website you can

  • go to get a lot of these things here.

  • And Katie went with Josh.

  • It's called whosedatedwho.com, and you get big graph,

  • you could start filling in the edges.

  • I don't know how reliable it is.

  • Now it's really critical that we're only looking edges

  • from here to here.

  • All right, so if there's an edge between Tom and Ben,

  • I don't want to know about it.

  • Just opposite-gender partners.

  • OK now in the USA, the number of nodes

  • here is about 300 million.

  • About three million people.

  • And the number of men nodes, male nodes, call these VM,

  • and this is VW, by the way, I'm using cardinality notation.

  • When I put bars around a set, that

  • is the denoting how many are in the set.

  • In the US there's about 147.6 men out of the 300.

  • And the number of women-- oh we got a w here--

  • is about 152.4 million.

  • So there's a little bit more nodes

  • on this side of the graph, than that side in the US.

  • What about the edges?

  • Any idea of how many edges there are here?

  • We don't know.

  • I sure as heck don't know how many edges there are.

  • So that we don't know.

  • The cardinality of the edge set we don't know,

  • and we're not likely to figure out.

  • I don't even think these surveys, really,

  • can estimate that.

  • But what we're trying to figure out

  • is the ratio of the average degree of the men,

  • to the average degree of the women.

  • Because the number of opposite-gender partners

  • you have is your degree here, and you're

  • looking for the average guy degree,

  • compared to the average female degree here.

  • That's what we're after.

  • All right so let's find that quantity.

  • Let's let A sub m equal the average number

  • of opposite-gender partners for men.

  • And we can let A W be the same thing for women.

  • All right.

  • Now we're trying to figure out the answer to this question.

  • What is A m, the average guy degree,

  • over the average woman degree.

  • And in particular, the University of Chicago

  • says, they say it's 1.74.

  • That the average guy as 74% more opposite-gender partners

  • than the average woman.

  • ABC News says it's 3.33, that is 233% more for the men,

  • than the women.

  • Now we're going to figure this out what this ratio is.

  • Just use a little bit of math here,

  • and a little bit of graph theory.

  • So let's write a formula for A m.

  • Well we're trying to figure out the average degree over here.

  • Well, that's pretty simple.

  • We just add up all the degrees, and divide

  • by the number of nodes.

  • And that'll give us the average degree.

  • So the average degree is the sum of the degrees,

  • over all men, x in the set of men,

  • of the degree of x, divided by the number of men.

  • Can somebody give me a simpler expression for this?

  • It doesn't have that nasty sum in it?

  • AUDIENCE: E.

  • PROFESSOR: E. The cardinality of E.

  • I'm adding all the degrees here.

  • Well that's just another way of counting all the edges,

  • because every edge shows up once, and only once,

  • in a degree count here.

  • And this is where, we use the fact we

  • have opposite-gender partners.

  • Because if I had some edges over here

  • they wouldn't get counted in sum of the degrees here.

  • All right so this is just the cardinality

  • of the number of edges, divided by the number of men.

  • Any questions about that?

  • Because this is an important statement about

  • graphs in general.

  • When I have a graph like this-- which

  • is called a bipartite graph, we'll

  • talk about more in a little bit.

  • But where the edges go from the left to the right

  • if I sum the degrees on the left,

  • I'm just counting the number of edges.

  • All right, let's figure out a formula for the average number

  • of partners for the women.

  • That simple that's just sum x over the women.

  • The degree of x, divided by the number of women.

  • Let me rewrite that so it's clearer.

  • What's a simpler expression for this?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yeah, this sum, adding

  • the degrees of the women, is just the number of edges,

  • right.

  • So that is cardinality of edges, divided by the number of women.

  • All right, well now we can write, solve for our formula,

  • average over men over average of the women.

  • That's E over VM, divided by E over VW.

  • Wow, this is nice.

  • I don't know the number of edges is, but it just canceled out.

  • And this is just the number of women,

  • divided by the number of men.

  • And in fact we know that.

  • That's this number, divided by that number,

  • which is about 1.0325.

  • So we just proved, that on average, a man

  • has 3%, or 3 and 1/4% more opposite-gender partners

  • than women.

  • No need to do the interviews, or spend years doing.

  • That is the answer.

  • And it has nothing to do with the promiscuity of men,

  • or women, nothing at all.

  • So the Chicago study is way off, and the ABC New study

  • is completely nuts.

  • It just can't be right, this is a proof.

  • Now what happened here?

  • Well what's going on, what's the reason for why this is true?

  • Yeah?

  • AUDIENCE: A male has a female partner then

  • the female has a male partner.

  • PROFESSOR: Yeah.

  • AUDIENCE: You're not looking at like how many males

  • are going to one female.

  • The promiscuity isn't even a part of the question.

  • PROFESSOR: That's right.

  • It takes two to tango.

  • Every time you got a guy, you got a women.

  • And you have the number of relationships going.

  • The average for the men is that number, divided by the men.

  • Average for the women is that same number,

  • divided by the women.

  • And so if there's more women, they're

  • going to have less partners on average.

  • Has to be.

  • So it really was a stupid question.

  • It's very, very simple to answer.

  • Now as it turns out there are endless studies like this,

  • in the literature.

  • In fact, a few years ago the Boston Globe

  • ran an explosive story about the study habits of students

  • on Boston-area campuses.

  • And their surveys show that, on average, minority students

  • tended to study with non-minority students

  • more than the other way around.

  • And they want on great length consulting the experts

  • as to why this might be true.

  • Why is it the minority students study

  • with non-minority students more than the other way around.

  • Now can anyone tell me why it is certainly

  • true, and not surprising, why that's the case?

  • AUDIENCE: Because they're the minority.

  • PROFESSOR: Because they're a minority.

  • There's fewer minorities than non-minorities.

  • End of story, we don't need this sociology PhD

  • from down the street to explain it to us.

  • We're going to see a lot of other bogus studies later.

  • This is not unusual, especially when we get the probability.

  • Just every day there's a new one in probability.

  • Any questions about this before we leave?

  • Unfortunately that's most all we'll say about sex today.

  • OK.

  • But now, in this example, we used an edge in the graph

  • to denote some kind of affinity between two nodes.

  • The two nodes liked each other in some sense

  • if they were connected by an edge,

  • or they had a relationship of some kind.

  • There's lots of examples in computer science

  • where you use an edge to denote just the opposite.

  • That the two nodes can't be near each other,

  • or don't like each other.

  • For example, consider the problem

  • of scheduling final exams at MIT.

  • And they do this after they find out all of your schedules,

  • and they try to schedule the exams so that you don't

  • have to take two at once, or there's as little

  • of that as possible.

  • For example, let's do an example here.

  • Say we look at these five classes.

  • Take 6041.

  • And this may not be totally accurate, but roughly.

  • So I've got five MIT classes, and I'm

  • going to put an edge between pairs of classes that have

  • overlapping student enrollment.

  • So in this case, for example, we've

  • assumed in the drawing of his graph,

  • that you can't have our exam the same time is 6002,

  • on the assumption there's students in both classes.

  • But you could have our exam the same time as 6034.

  • Because there's not an overlapping student

  • in both classes, so the exams could

  • be scheduled at the same time.

  • So we've used a graph to represent

  • which courses can't have their exam at the same time.

  • Now let's also suppose we have a set of slots for the exam.

  • And say they're all on a Wednesday.

  • And the first slot is Wednesday from 5:00 to 7:00.

  • And the next one is 7:00 to 9:00.

  • And then, the next one is 9:00 to 11:00.

  • And then 11:00 to 1:00 in the morning, and then 1:00 to 3:00,

  • getting pretty late.

  • And your job is to figure out how not to have

  • to use these later exam slots.

  • You'd like to use as few as possible

  • so you're not going too late night,

  • or come before the holidays, so you're not

  • having exams on Christmas and New Year's, for example.

  • So the goal is to assign slots to the nodes.

  • Put every node in a slot so you don't

  • have nodes hooked by an edge getting the same slot.

  • Now this is an example of what's called

  • a graph coloring problem.

  • So let's define that.

  • Given a graph G, and K colors, assign a color to each node,

  • so that adjacent nodes get different colors.

  • All right, and then the minimum number of colors you need

  • is called the chromatic number of the graph.

  • So the minimum value of K, for which such a coloring exist,

  • is the chromatic number OF the graph.

  • And it's denoted by this symbol chi of G.

  • Because usually you want to use a small number of colors.

  • Now what does a color represent when we're

  • dealing with this problem?

  • What's the meaning of a color?

  • AUDIENCE: Time slot.

  • PROFESSOR: A time slot, OK.

  • So let's call this time slot C1, C2, C3, C4, C5, so there's

  • five possible colors.

  • Now of course, we could color this graph with five colors,

  • every node could just get its own color.

  • But then somebody's taking their exam from 1:00 to 3:00 AM,

  • and that's a bit of a pain.

  • Let's see if we can do less than five.

  • Let's say I give this color one, let's give this one color

  • one, that's OK, because they're not connected.

  • I can't give this one color one, so I give it color two, say.

  • Now this one I can't give color one, because this guy got it,

  • he can't get color two, because that guy got it.

  • So it give it color three.

  • And well, I can't do one, two, or three here,

  • so I gotta go to color four.

  • All right so 6042 will get the 11:00 PM to 1:00 AM slot,

  • not so good.

  • Can we do any better?

  • Can we get away with three colors.

  • Some say yes, some say no.

  • How many people think you can do three colors on this graph?

  • A bunch.

  • How many think you can't do any better?

  • All right, the vote is mostly for three.

  • Let's see.

  • Any ideas?

  • Anybody see how to do three?

  • Yeah?

  • AUDIENCE: Assign C4 to 6034 .

  • PROFESSOR: Assign C4 to 6043.

  • AUDIENCE: Or C1 to 6042.

  • PROFESSOR: C-- I can't do see C1 to 6042.

  • It crashes, but can I do-- yeah?

  • Put

  • AUDIENCE: C1 in 6003.

  • PROFESSOR: C1 in 6003.

  • AUDIENCE: And get rid of C1 in 6034.

  • PROFESSOR: Get rid of--

  • AUDIENCE: Make it C2.

  • PROFESSOR: Make this a C2.

  • Oh, yeah.

  • All right, these got C1, they're not adjacent.

  • These got C2, they're not adjacent.

  • This can now get C3.

  • So we can have our exam from 9:00 to 11:00, which is better.

  • All right, can anybody do it in two colors?

  • Can anybody offer a reason why two colors may not be possible?

  • Yeah?

  • AUDIENCE: Because let's say you could do it with two colors.

  • PROFESSOR: Yep.

  • AUDIENCE: 6041 and 6002 have to be different colors.

  • PROFESSOR: Yes.

  • AUDIENCE: 6042 can't be C1, and it can't be C2.

  • PROFESSOR: Yeah, good.

  • So you can't in two colors, because these three guys

  • would violate that.

  • You've got a triangle here.

  • Each one of these guys has to be different than the other two.

  • So two colors can't work.

  • You've got to have at least three in this case.

  • So three is optimal.

  • We have just shown for this graph,

  • the chromatic number is three.

  • All right, now in general doing what we just did is very hard.

  • No one knows a fast algorithm for determining

  • the chromatic number.

  • In fact, it's a weird kind of problem,

  • because it's easy enough to check that a coloring is OK.

  • If somebody put a coloring on the board, you can check,

  • oh that works really simply.

  • Just check every edge, and make sure the colors are different.

  • But figuring it out, as best we know,

  • you've got to try an exponential number of possibilities.

  • So if I had 100 nodes here, my running time

  • of the algorithm to check all the possibilities

  • would be exponential and a hundred.

  • Yeah?

  • AUDIENCE: Can that number just like the highest

  • degree of each node, or nodes.

  • PROFESSOR: Uh no.

  • But it's no worse than something like that,

  • as we'll see a few minutes.

  • That's a great observation.

  • And we're going to come back to that in a few minutes.

  • But it's not just that.

  • OK now in fact even figuring out for an arbitrary graph

  • if three colors can be done, called

  • the three-coloring problem, that's really hard.

  • No one knows how to solve that in less than exponential time.

  • In fact, one of these NP-complete problems

  • is what it's called.

  • How many people here don't know about NP-completeness?

  • Is everybody-- all right so all of you

  • haven't seen NP-completeness.

  • OK so there is a class of thousands of problems--

  • in fact there's books list these 1,000 problems-- that are all

  • NP-complete, somebody's proved they belong in the class.

  • And what that means is that if somebody gave you

  • a solution, like a coloring here,

  • it's easy to check really quickly if it's valid.

  • But figuring it out is really hard.

  • And if you figured out how to solve

  • one of those thousands of problems,

  • like suddenly you figured out how to tell if any graph could

  • work with three colors, you would solve automatically

  • all other thousands in the book.

  • So it's this book of problems you will constantly

  • run into in your career in computer science.

  • And it's bad when you run into one,

  • because there's no good algorithm to solve it known.

  • But if you just solved one of them,

  • the other thousands would suddenly be solvable quickly.

  • Even better, you win a million dollar prize.

  • One of these Millennium Prizes we

  • talked about the first lecture.

  • Even if you show you can't find a fast algorithm for one

  • of them, that means that known of them have fast algorithms,

  • and you also get a million dollars.

  • So this is the central problem in computer science, and theory

  • computing, is whether or not you could solve

  • these NP-complete problems.

  • Now actually lots of people have claim to do it.

  • And in fact, there was a lot of buzz in the community

  • about a month ago when actually a reputable researcher

  • at HP Labs said he'd done it.

  • He proved that you can't solve NP-complete problems.

  • And he got people going for probably at least a week,

  • until they discovered a fatal flaw.

  • And the proof was actually bogus.

  • So no one still knows if you can solve

  • these NP-complete problems quickly.

  • Now the problem is, in practice, you run into these things

  • all the time, like MIT really does

  • have to schedule the exams.

  • So you've got to do something.

  • You can't just go say, hey it's NP-complete, so no exams

  • this year, or whatever.

  • That's not going to fly, so you got to do something.

  • So now this is a problem-- many of you

  • when you go into careers, you're going to be faced with this.

  • You got to do something.

  • Any thoughts about an algorithm for coloring graphs that might

  • use a small number of colors?

  • It doesn't have to always work, or you're

  • going to win a lot of money if it does.

  • But a simple algorithm, you can't

  • take either the 100 steps.

  • You got to be linear, probably, or quadratic time.

  • That could get you a small number of colors.

  • Any thoughts about what you'd do?

  • Yeah?

  • AUDIENCE: The number of degrees and nodes?

  • PROFESSOR: The number-- what about it?

  • AUDIENCE: The highest degree and that node,

  • the 6042 is [INAUDIBLE].

  • PROFESSOR: Yeah.

  • AUDIENCE: So you could use that.

  • PROFESSOR: Good, all right.

  • So what do I do with that-- so I found

  • a node with a high degree, there's

  • three of them have degree three here.

  • What do I do with them?

  • AUDIENCE: Pick a different color to.

  • PROFESSOR: Pick a different color,

  • that means I've colored some of the others.

  • If I pick a different color, do I start with them,

  • or do I finish with a high degree nodes?

  • Because you've got to assign the colors to them.

  • And high degree is important to be thinking about.

  • We're going to prove a theorem in just a minute about related

  • to degree and coloring.

  • AUDIENCE: Start with them.

  • PROFESSOR: Start with them, and do what with it?

  • Color?

  • AUDIENCE: Yeah, and then assign the ones that aren't connected

  • [INAUDIBLE] to the same slots.

  • PROFESSOR: OK, so I could-- here's a degree of theory

  • now I can start with color one for that.

  • And then what do I do next?

  • I pick-- its neighbors have to get different colors, I guess.

  • You'd start coloring the neighbors.

  • AUDIENCE: My first instinct would be

  • to color all the [INAUDIBLE].

  • PROFESSOR: OK.

  • And what color would use for them?

  • AUDIENCE: Different ones.

  • PROFESSOR: Different ones if they're connected,

  • or if they're not connected you'd still use different ones?

  • AUDIENCE: Only if they're connected.

  • PROFESSOR: Only they're connected use different ones.

  • And so if they're not connected, you'd use the same colors?

  • Yeah?

  • You're going close, and it actually works pretty well.

  • The underlying principle you're sort of thinking about here

  • is you've got some notion of the order in which you're

  • going to process your graph.

  • And you're going to start with a high degree nodes,

  • in your case.

  • And as you go along, you're going

  • to start coloring the nodes.

  • And you're going to make sure you color them legally.

  • And it sounds like you're going to color them with a low color

  • as you go along.

  • And that is probably the most basic graph coloring approach.

  • And almost you could almost say is a generic approach.

  • So let's define that, and then see prove some facts about it.

  • Most of the graph coloring algorithms in practice

  • are based on this approach.

  • And we're going to call it the basic graph coloring algorithm.

  • And for our graph G, with vertices V, and edges E.

  • So the first step is going to be to order the nodes from 1 to n.

  • Now in your case, you were suggesting

  • an ordering where I have the high degree nodes first.

  • All right.

  • But for now we're not going to specify that.

  • We're going to make it any ordering you want.

  • And then we're going to have a notion of an order

  • on the colors, as well.

  • And I don't know how many colors,

  • but they're going to be numbered 1, 2, and so forth.

  • And then we're going to process the nodes one at a time,

  • to N. We color the nodes, what is step I,

  • we color the Ith node V sub i with the lowest legal color.

  • And by the legal I mean you don't color at the same node

  • as another node that's already been colored the same that it's

  • adjacent to.

  • All right so let's try this.

  • In fact, this is sort of the algorithm I used initially

  • to color exam graph over there.

  • All right, so let's look at that.

  • So let's say we-- let me erase the colors here, and put

  • an ordering on the nodes.

  • So let's say I ordered them with 6034 first,

  • so this would be V1.

  • Then 6041 is V2.

  • Then V3, V4, V5.

  • If that's my ordering, what color would I assign to 6034?

  • AUDIENCE: One.

  • PROFESSOR: One, C1, I'd color it first to get C1.

  • What color does 6041 get?

  • C1, as well, it's the lowest possible color

  • that's legal, and is not hooked to this guy, so C1 is legal.

  • What color do I give here?

  • C2.

  • Then I color this one next C-- can't do C2, can't do C1,

  • so I pick C3.

  • And then I get to 6042 last, and I

  • can't do one, two, or three, so I do four.

  • All right so algorithm, with that ordering,

  • gave four colors.

  • However we know there's a way to do a different ordering that

  • gives us three colors.

  • In particular, let's see if we do

  • this what happens if we use this other ordering.

  • Let me erase these.

  • Say that's V1, V2, V3, V4, V5.

  • Now I get C1, this will be C2, C1.

  • What's this one get?

  • C2.

  • Ah, much better.

  • C3.

  • So different orderings result in different numbers of colors

  • here.

  • So the whole art now becomes finding a clever ordering.

  • And so many people have already had good ideas,

  • pick the largest degree nodes first.

  • And in fact, if you simulate the algorithm on lots of graphs,

  • you do better on average when you color the larger degree

  • nodes first.

  • And then if you start to use more exotic orderings,

  • you can do even better.

  • If you take a lot of graphs that are out there,

  • and run your algorithm, and see how well you do,

  • you do better with more sophisticated orderings.

  • In fact, this was my senior thesis back

  • when I was undergraduate student.

  • I was trying to figure out better and better orderings

  • that worked for graphs.

  • And at the time it caused a bit of a problem.

  • I was a undergraduate at Princeton.

  • And Princeton, to this day I think,

  • still has exams after the holidays, the Christmas

  • holidays, New Year's holidays.

  • And the students wanted to have the exams before Christmas,

  • because they hated going home for the holiday,

  • and then you've got to worry about your exams

  • when you come back.

  • And the faculty said no, there's no way

  • to get them all compressed into a small number of days.

  • Now I wasn't aware of all that of the time.

  • But my thesis was go figure out good ordering.

  • So I tried lots of different orderings.

  • And I tried the largest degree first,

  • and recursive versions of that actually worked very well.

  • And then tried it on the Princeton exam graph.

  • And lo and behold, you could actually squish it down,

  • so you could give all the exams, I think was,

  • 4 and 1/2 days, plenty of time to give them before Christmas.

  • Which caused a fair of scandal at the time,

  • because then the faculty had to come clean

  • that they just didn't want to bother having

  • the exams before Christmas.

  • Now this algorithm is an example of what's

  • known as a greedy algorithm.

  • Now in a greedy algorithm it's always simple.

  • You just go one step after the next,

  • taking the best you can do at each stop.

  • You never go back and try to make things better.

  • You never do hill climbing, if you're familiar with that term.

  • You just always keep it simple, one thing after the next,

  • very fast.

  • Sometimes it works great in practice.

  • Sometimes it doesn't.

  • But it's always where you start, some simple approach like this.

  • Now this algorithm actually, even

  • if you don't try to monkey with the ordering,

  • even for a worst case ordering of the nodes,

  • that actually does pretty good for a lot of graphs.

  • And in fact, it does really well--

  • as somebody already asked about--

  • if all the nodes have low degree.

  • So let's state that as a theorem.

  • And then we're going to prove that.

  • So if every node in a graph G has degree, at most,

  • d-- so that's the biggest degree in the graph, D-- then

  • this basic algorithm uses, at most, d plus 1 colors for G.

  • No matter what the ordering is, you'll

  • never do worse than d plus 1 colors.

  • So what's the value of d for our exam graph over here?

  • d is 3.

  • Every node has degree, at most, three.

  • And so it says, that no matter what ordering you picked here,

  • you'd get at most four colors.

  • Now you might do better.

  • In fact, we found an ordering that got three.

  • So it's possible to do better.

  • So let's prove this fact because this makes a difference.

  • Say you have a graph with hundreds of nodes.

  • But every node has degree, at most, three.

  • Well that says you only need four colors even,

  • if the graph has 1,000 nodes, and that's very useful.

  • So in that kind of situation it does very well.

  • So let's prove that.

  • Any ideas as to what proof technique we're going to use?

  • AUDIENCE: Invariant.

  • PROFESSOR: Invariant, close.

  • Not quite an invariant, but close.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: What?

  • AUDIENCE: Well ordering principle.

  • PROFESSOR: You know well ordering principle, yeah,

  • we're going to use the equivalent version of that.

  • We're going to use induction.

  • If you like well-- it's equivalent to well ordering.

  • If you like well ordering you could do it that way.

  • I think it's easier using induction here.

  • So the proof is by induction.

  • All right so the first thing we need

  • is an induction hypothesis.

  • Any thoughts about what the induction hypothesis should be?

  • Yeah?

  • AUDIENCE: If you have a graph with n nodes then

  • where the degree of any nodes is less than [INAUDIBLE]

  • then you can do it.

  • PROFESSOR: That's great.

  • You're going to do really well on the midterm,

  • because you put an n into this thing,

  • but there's not an n here to start.

  • What are most people going to do--

  • we used to ask this actually.

  • We asked this once on a test many years ago,

  • and it was an utter disaster, because did everybody do?

  • May be one student, or two, put an n into there.

  • But what's the naturally thing to do to induct on here when

  • you look at this statement?

  • You're going to induct on d, because the first thing you do

  • is you make this be your induction hypothesis.

  • There's only one thing to use, so you're

  • going to have your predicate be p of d,

  • and it's going to be that.

  • Now It didn't occur to us that's what everybody was going to do,

  • but it should have.

  • They all did that and it was a disaster.

  • Because if you do this, well you've

  • got to take a graph with maximum degree d, or d plus 1

  • in the inductive step, pull out all the nodes

  • with degree d plus 1 to get a graph with now degree d.

  • And that's a mess.

  • You just pulled out a lot of nodes, potentially.

  • Color that in d plus 1 colors, now put all that junk back in.

  • And say only used one more color.

  • Nightmare.

  • And these were MIT students under pressure.

  • It was a nightmare.

  • So that does not work.

  • And in fact, we will ask an induction question

  • on graphs on every test you take in this course.

  • It will happen.

  • And so usually, with induction, you

  • take this as your induction hypothesis.

  • With graphs, you have to be careful.

  • And worst part about this is we tell people

  • when this doesn't work, use a stronger induction hypothesis.

  • So students tried to make a stronger,

  • but they're still stuck on d, and it was still a disaster.

  • With graphs, you do something different.

  • And the first thing you do with a graph, usually,

  • is put n in here.

  • And if it doesn't work with n, the number of nodes,

  • you put in e the number of edges.

  • And induct on that.

  • And so what you said is exactly the right thing to do.

  • Don't do this, or least don't spend too much time on it.

  • Pretty quickly try this.

  • If every end node graph-- if every node in an n

  • node graph G has degree at most degree,

  • then the basic algorithm uses at most d, plus one colors.

  • And now you induct on n.

  • And almost always on graphs, that's the first thing to try.

  • Even if it's not in your theorem statement.

  • Any questions about that?

  • Well let's start with this, and see

  • if we can make this one work.

  • So what's the next step in our proof?

  • What do we got to do?

  • Base case.

  • And the base case will be, not n equals 0,

  • because we can't have a zero node graph, but n equals 1.

  • And how many edges do we have?

  • Zero.

  • If there's one node, we don't allow loops,

  • so it's zero edges, which means that the degree of our graph

  • has to be zero.

  • There's no edges.

  • And of course there's only one node,

  • so one color is going to work, and that

  • happens to equal d plus 1.

  • All right, so the base case is true.

  • For one node graphs, you can always

  • use d plus 1 colors, where d is the max degree.

  • All right, next we have the inductive step.

  • So here we assume P n is true for the induction.

  • And now we look at an n plus 1 node graph

  • to show P n plus 1 is true.

  • So we let G be any N plus 1 node graph.

  • We got to show you can color it in d plus 1 colors.

  • And let's let d be the max degree, the largest

  • degree in G.

  • We've got to show we can color it in d plus 1 colors.

  • Well the basic algorithm, let's say.

  • First thing we do is we order the nodes

  • in an arbitrary order.

  • And we're going to show whatever order you pick is OK.

  • All right so what are the nodes?

  • Anyway at all.

  • Now how am I going to use the induction hypothesis?

  • I know, I can assume, the for any N node graph

  • I can color it in the max degree plus 1 colors.

  • How am I going to use that to help me color G here,

  • the n plus 1 node graph?

  • Any thoughts?

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yeah, let's create an n node graph

  • by looking at these nodes, and taking

  • this one out of the time being.

  • Remove the last V n plus 1 node in the order.

  • That leaves an n node graph.

  • So let's write that down.

  • We remove the n plus 1 from G. And that creates a new graph,

  • call it G prime with vertices, V prime and edges, E prime.

  • So we create a new graph by removing that node.

  • And we remove all the edges tied to that node.

  • So for example over here, the last node

  • was 6042, so we take out 6042, and all these edges.

  • And this is a graph that we're left with.

  • That graph has n nodes.

  • What's the maximum degree in G prime?

  • When I pull out a node, can the degree of any node go up?

  • No, I'm just taking stuff out.

  • So I know that G prime has maximum degree, at most, d.

  • The degree didn't go up of any node.

  • Might have gone down, but it didn't go up.

  • So G prime has max degree, at most, d, and it has n nodes.

  • So we can use the induction hypothesis P n.

  • It says that the basic algorithm uses d plus 1,

  • at most, d plus 1 colors for nodes V1 to V n.

  • Any questions about that?

  • So if this were the n plus first node, last node in the ordering

  • take it out.

  • The basic algorithm now, take the same order here,

  • V1, V2, V3, V4, basic, we'll color that in d plus 1 colors.

  • And all I have left is to give this guy color,

  • and I'll have color G. Question?

  • No.

  • All right.

  • So by induction I've colored these guys, V1 to V2,

  • and d plus 1 colors, all that I have left to do

  • is color V n plus 1.

  • And hopefully we're not going to use color d plus 2,

  • because then we sort of-- it wouldn't work.

  • We got to use one of the first d plus 1.

  • All right, so let's look at V n plus 1.

  • And let's call its neighbors in G, U1, U2, Ud.

  • It has, at most d neighbors, because every node in G has,

  • at most, degree d.

  • A neighbor's a node you're adjacent to.

  • All right so, V n plus 1 has at most d neighbors, is adjacent

  • to, at most, d other nodes.

  • Now what does that mean about the color I

  • can use on V n plus 1?

  • What do I know about what color I can use for that?

  • Yeah?

  • AUDIENCE: It can't be any of the colors of U1, U2, and so on.

  • PROFESSOR: It can't be any one of these colors that

  • were assigned here.

  • That's true.

  • So how many colors got ruled out?

  • At most d, and how many am I working with?

  • d Plus 1.

  • So I got one left that I can use safely.

  • OK.

  • So this means there exists at least one color

  • in my set of d plus 1 colors.

  • It's not used by any neighbor.

  • And we're going to give V n plus 1 that color.

  • All right.

  • So now I've colored every node in G, the n plus 1 node graph,

  • safely using a total of d plus 1 colors.

  • So that means the basic algorithm uses,

  • at most, d plus 1 colors, on G. That means P n plus 1

  • is true-- whoops-- and the induction is complete.

  • Any questions?

  • Yeah.

  • AUDIENCE: Could you also start from the other way,

  • and start 1, go to 2 nodes, 3 nodes at each step keeping

  • all nodes at all other nodes.

  • [INAUDIBLE]

  • PROFESSOR: What do you mean by keeping all nodes connected?

  • AUDIENCE: [INAUDIBLE] each node has an edge connecting

  • to each other one.

  • PROFESSOR: OK so, then I get a specific graph.

  • I start with this, I add a node and make it adjacent.

  • I add a node and make it adjacent.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yeah.

  • So you've constructed a particular graph.

  • This is actually called, for the n nodes, it's called Kn,

  • is the n node complete graph, also called a clique,

  • like a clique of friends, where everybody likes everybody,

  • in a clique.

  • And in fact for n here, for those n nodes,

  • what's the max degree?

  • Max degree is n minus 1.

  • What's the chromatic number of this graph?

  • What's the minimum number of colors?

  • [INTERPOSING VOICES]

  • PROFESSOR: And they all have to be different,

  • which is d plus 1.

  • So you have built a special graph

  • for which the optimum of number colors is d plus 1.

  • But that is not a proof that this is true for all graphs.

  • Because you've looked at a particular graph here.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: What's that?

  • AUDIENCE: [INAUDIBLE] It means that you can still use your

  • less than or equal to sign.

  • PROFESSOR: I see, so you'd add a node,

  • and it's only connected to a few of them.

  • AUDIENCE: No, it's connected to all of them,

  • but it still implies that you need less than

  • or equal to the colors.

  • It turns out it happens to be equal to.

  • PROFESSOR: Yes, in this case that's right.

  • So you've made an argument for this case

  • where it actually is equal, but that

  • only worked for this graph.

  • AUDIENCE: [INAUDIBLE] worse case.

  • PROFESSOR: It is the worst case, so it meets the bound.

  • It shows you cannot improve this bound.

  • Yeah, is there a question up there?

  • AUDIENCE: All I was going to say is that you've

  • proved it's the worst case.

  • PROFESSOR: Right, so what you've done here

  • is you've shown that I could not make that theorem any stronger.

  • I could not replace it with d here.

  • All right.

  • Because you've given an example where

  • I can't get d colors, where the maximum degree is d.

  • But that doesn't-- To get a proof for a theorem,

  • I got to go through all this.

  • That wouldn't give me a proof of the theorem.

  • They're not equivalent.

  • One's an upper bound, one's an existence of a lower bound.

  • This shows that for any graph, you need at most d plus 1.

  • So any graph, at most.

  • That shows there is a graph that you need at least.

  • And they are not equivalent.

  • All right.

  • One is for all, and upper bound.

  • The other is there exists a lower bound.

  • So different in two ways that are important.

  • This kind of proof is very typical for what you'll

  • see with induction in graphs.

  • And you'll get a lot of practice with it.

  • Are there any other questions on this proof?

  • OK.

  • All right, see we've seen now, by that example,

  • we can't improve the theorem.

  • In some cases, though, the theorem

  • is way off, for some graphs.

  • Can anybody think of a graph where

  • the bound we get from the theorem, of d plus 1 colors,

  • is way off from the actual chromatic number you need,

  • the number of colors you need?

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: What is it?

  • AUDIENCE: A graph [INAUDIBLE] two sets of [INAUDIBLE]

  • PROFESSOR: Good, OK.

  • Yes, so what if we did this graph.

  • Let me draw it out.

  • So you've got a bunch of nodes here, bunch of nodes here.

  • And every node here is connected to every node

  • over the other side.

  • And if this is an n no graph, and I've got n over 2

  • on each side, what's my degree here?

  • What's my max degree of this graph?

  • AUDIENCE: N over 2.

  • PROFESSOR: N over 2.

  • So d is n over 2.

  • What's the chromatic number?

  • How many colors do I need for this?

  • Two.

  • All right, so d plus 1 is way off of two.

  • There is a even worse example.

  • Yeah?

  • AUDIENCE: That graph where you have one node center that's

  • connected to a bunch of nodes regularly distributed about.

  • PROFESSOR: Yeah, the star graph.

  • All right, so I got one of the center, I got n minus 1

  • outside.

  • So here the maximum degree is n minus 1,

  • just like a complete graph.

  • But how many colors do I need?

  • Two.

  • So it's even worse here.

  • All right now what about the basic algorithm?

  • How well does the basic algorithm do on this graph?

  • Or to the vertices some way?

  • Color on one [INAUDIBLE] lowest color.

  • How many colors is it going to use?

  • AUDIENCE: Two.

  • PROFESSOR: Two.

  • It doesn't matter the vertices.

  • V1, V2, V3, V4, because I'll color this one 1.

  • What am I going to call that one?

  • 1.

  • Then I get to the center, what am I going to color it?

  • 2.

  • And now all the arms, what do they get colored?

  • They all get 1.

  • Whatever order you pick, you get two colors.

  • All right so now there's a difference

  • between the theorem just gives you an upper bound, it says,

  • at most, d plus 1 colors.

  • But in fact the algorithm can do a lot better

  • than that, as on this example.

  • So the algorithm might be a lot better.

  • Everybody see that what we're doing here?

  • How the algorithm is better than the bound

  • we proved by the theorem, even though the bound was

  • pretty good for some graphs.

  • Now it turns out-- I mean we're not

  • going to win a million dollars for this algorithm.

  • And in fact, this algorithm is sometimes very bad.

  • And a really bad example it's very close to this.

  • In fact actually this one, let's look at how well does basic

  • do one this one here.

  • Make some ordering.

  • V1, V2, V3.

  • What's the basic algorithm going to do on this complete--

  • it's called a complete bipartite graph, is what's this called.

  • I'll define bipartite in a minute--

  • but what's the basic algorithm do here?

  • Any idea-- does it take n over 2 colors, or does it take 2?

  • Any ideas?

  • 2.

  • So take a vertex, and the first one, say V1s here, get C1.

  • As long as I keep picking vertices over on this side,

  • they're going to get C1.

  • As soon as I get to a vertex over here,

  • what color does it have to get?

  • AUDIENCE: C2.

  • PROFESSOR: C2 because it's touching

  • the very first one we had here.

  • So when I get vertices over here,

  • they're all going to be C2.

  • When I go back over here, they're going to be back to C1.

  • So actually basic does good here too, gives you two colors.

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Ah, those two aren't connected.

  • But this case, if I've got a vertex over here

  • it is, by definition, connected to the vertex over here.

  • Because every possible edge is here.

  • But that's a great idea.

  • What if they weren't all connected,

  • that's actually a great idea.

  • In fact, the nasty example for the basic algorithm

  • is very much like that.

  • Let's draw it.

  • Because so far, the basic algorithm

  • is pretty much done perfectly on all the graphs

  • we looked at even when the theorem wasn't tight.

  • So here is a nasty graph.

  • And it is very close to the graph we just look like,

  • where all the edges are there.

  • In this case, all the edges are there,

  • except for the one straight across.

  • So if this is-- the edge denotes likes,

  • this is a world where you like everybody but your spouse.

  • All right, so you have an edge to every one,

  • except the one directly across from you.

  • No edge there, and so forth.

  • So it has almost every edge, but it's missing these edges.

  • Now the basic algorithm might do well here.

  • What would be a good ordering for this graph

  • to label these V1 through Vn?

  • Yeah?

  • AUDIENCE: Go through everything on the left side,

  • and then the right side.

  • PROFESSOR: Yeah, that's right.

  • Because then color 1, color 1, color 1, all the way down.

  • One color for the left, what does this one get?

  • Color 2, because it's hooked up against.

  • And these all get color 2, so I've used two colors.

  • Really good.

  • Basic algorithm's looking great.

  • Now here's a harder question.

  • Can you figure out a bad ordering

  • for this graph, where I use a lot more than two colors.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: What is it?

  • AUDIENCE: It starts at the top of the cross, and then

  • the next level then across.

  • PROFESSOR: Very good.

  • V1, V2.

  • Just as natural, really, if think about it,

  • to order it this way.

  • All right.

  • What color does V1 get?

  • C1.

  • What color does V2 get?

  • AUDIENCE: C1.

  • PROFESSOR: C1 because it's not hooked up here.

  • What color does V3 get?

  • AUDIENCE: C2.

  • PROFESSOR: C2.

  • What about V4?

  • AUDIENCE: C2.

  • PROFESSOR: C2.

  • It's not hooked up.

  • It can't get one, because that's up here.

  • And it's not the two, so it gets two What color does V5 get?

  • AUDIENCE: C3.

  • PROFESSOR: C3.

  • Because it's hooked up to one to two.

  • V6 ?

  • AUDIENCE: C3.

  • PROFESSOR: C3, it's hooked up to one and two, but not three.

  • And you can see what's happening here.

  • All the way down here he's hooked up

  • to all the n over 2 minus 1 colors.

  • So he also takes C n over 2.

  • So if you pick that ordering, not so good.

  • You use n over two colors.

  • So it really matters the ordering.

  • Now I should say graphs like-- actually any questions

  • about what we did here?

  • About this?

  • All right, now I should say that graphs like this

  • have a special name, they're called bipartite graphs.

  • And that's important to remember.

  • All right, so a graph G is said to be bipartite

  • if the vertices can be split into two sets, or partitioned,

  • and we'll call them a left set, and a right set,

  • so that all the edges connect a node in the left set,

  • to a node in the right set.

  • So in fact, a lot of today we've been

  • looking at bipartite graphs, because the nodes are here.

  • Like the men, and the women, and the edges only go from the left

  • to the right.

  • And that is called bipartite.

  • And it's called bipartite because you

  • can do it with two colors, or in two pieces.

  • So you don't win a million dollars

  • for deciding whether or not a graph can

  • be colored in two colors.

  • That's easy.

  • You'll even do it for homework one of these times.

  • You do win the million dollars for deciding if a graph can

  • be colored in three colors.

  • That's really hard to do.

  • Now coloring problems come up in all sorts of applications.

  • You know with this company, Akamai, that came out of MIT,

  • we've talked about.

  • We run a network of 75,000 servers.

  • And they're used to distribute content on the internet,

  • and so forth.

  • And we have to deploy a new version

  • of our software on those servers,

  • pretty much every week.

  • We're pushing new software out.

  • And you can't deploy on every server at the same time,

  • because you've got to take down a server

  • to deploy new software on it.

  • Got to take it out of commission.

  • And so we can't just take down all 75,000 servers,

  • because then all the Facebook, and Netflix,

  • and all those sites would stop.

  • That would be bad.

  • And we can't do them one at a time, because there's 75,000.

  • And it takes a few hours for each one

  • to get the traffic off, stop it, load new software,

  • and turn it back on.

  • And it would take us years to do one software install,

  • which we got to do every week.

  • So we've got to figure out a schedule for how many servers

  • you take down at a given time, and which ones.

  • And it turns out pairs of servers

  • have certain critical functions.

  • So there's certain pairs of servers you can't take down

  • at the same time.

  • So we have a gigantic 75,000 node coloring problem,

  • where there's edges between servers.

  • Nodes are servers, and there's an edge

  • between if you can't install new software at the same time.

  • And so when it turns out, when you

  • run one of these graph coloring algorithms on it,

  • you could do it with eight colors.

  • It just turns out that way.

  • So that means there's eight waves of install

  • that go on to the network.

  • And now eight times a few hours each

  • means that we can do it in a day, and you can manage it.

  • You know on a much smaller scale,

  • the same problem exists for register allocation,

  • for variables.

  • Here you've got to assign every variable to register.

  • But you can't have variables that

  • are active at the same time associated

  • with the same register.

  • And you want to minimize the number of registers you need.

  • So again, you have the graph coloring problem.

  • The number of colors is the number of registers you need.

  • And two variables can't get the same color if their active

  • at the same time, so you put an edge between them.

  • The most famous example of graph coloring

  • is the map coloring problem, with the four coloring theorem.

  • And so here, every country is a node.

  • Adjacent countries have an edge between them,

  • because you don't want to color adjacent countries

  • the same color, or you can't tell

  • they're different countries.

  • Now the last example we can talk about

  • is an important problem in communication theory,

  • communication networks, where again coloring comes up.

  • Now here you need to assign frequencies to radio stations,

  • or the cell towers.

  • It comes up in mobile networks, or just in with radio stations.

  • And if two towers have an overlapping area,

  • they can't be given the same frequency,

  • so you get collisions between the towers.

  • And frequencies are very expensive.

  • Companies pay the government a lot of money

  • to get certain spectrum.

  • So suppose you had this problem.

  • Here's tower A, this is A's range, where it reaches.

  • Here's tower B, so it overlaps some

  • with A. Here's tower C. Here's tower E. And here's tower D.

  • All right now the question would be,

  • how many radio frequencies do you need?

  • What's the minimum number of frequencies you need

  • to enable all the towers here?

  • We could make that be a graph.

  • There's a node for each tower.

  • And an edge between towers, if they overlap.

  • C doesn't overlap with B, E does.

  • E overlaps here.

  • And then D overlaps here.

  • So how many frequencies do you need for this graph?

  • AUDIENCE: Four.

  • PROFESSOR: Four would work, three is better.

  • Can you do two?

  • No you can't do two, because you got here.

  • But you could do three.

  • You could do one, two, three, two, one.

  • This problem comes up--

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Did I screw up?

  • Ooh, no I can't do that.

  • One, two, yeah much better.

  • All right, this problem comes up all over the place.

  • I'm certain you'll see it sometime in your career,

  • you'll have some problem, or you're scheduling something,

  • and it's really a graph problem in disguise.

  • OK that's it for today.

The following content is provided under a Creative

字幕與單字

單字即點即查 點擊單字可以查詢單字解釋

B1 中級 美國腔

Lec 6 | 麻省理工學院 6.042J 計算機科學數學,2010年秋季。 (Lec 6 | MIT 6.042J Mathematics for Computer Science, Fall 2010)

  • 79 5
    zero2005x 發佈於 2021 年 01 月 14 日
影片單字