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  • PROFESSOR: OK, on Tuesday we talked about sex,

  • so today we're going to talk about marriage.

  • Now, in terms of graph theory, marriage

  • is expressed as a matching problem,

  • and today we're going to talk about a matching algorithm that

  • is used in all sorts of applications.

  • It's used by online dating agencies

  • to match compatible people together.

  • It's used for assignment problems,

  • for example, matching interns to hospitals on match day.

  • It's used for resource allocation problems,

  • for example, load balancing traffic on the internet.

  • And we'll talk about the applications

  • at the end of class.

  • In its simplest form of a matching problem,

  • you have a graph where the edges represent compatibility.

  • Two nodes can be paired together, or married,

  • and the goal is to create the maximum number

  • of compatible pairs.

  • So let's define a matching, given a graph, G,

  • with nodes, V, and edges, E. In matching,

  • you can think of it as a collection of edges,

  • or a subgraph of G where every node has degree 1.

  • So everybody can be married just to one person.

  • So let's draw an example, maybe not put that edge in.

  • And let's label these nodes x1, x2, x3, x4, x5, x6, x7, and x8.

  • Now x1, x6 and x2, x5 is a matching, so x1, x6 and x2,

  • x5 is a matching with two edges, so we say it has size two.

  • All right, so I can pair these guys up and pair these guys up.

  • Is there any bigger matching in this graph?

  • So I found one with two marriages, here

  • and here, two edges.

  • Yeah?

  • AUDIENCE: x1, x7.

  • PROFESSOR: x1, x7.

  • AUDIENCE: x2, x6.

  • PROFESSOR: x2, x6.

  • AUDIENCE: [INAUDIBLE].

  • PROFESSOR: Good.

  • All right, that's a matching of size three.

  • So I got three couples together.

  • Good.

  • Can I make a bigger matching, one with size four--

  • four marriages here?

  • No.

  • Why not?

  • Can anybody give me a reason why it can't be done?

  • Yeah?

  • AUDIENCE: x8 and x7 would have to be matched with someone.

  • PROFESSOR: Yeah, so if-- Yeah?

  • AUDIENCE: They could only be paired with x1,

  • but x1 can't be paired with both.

  • PROFESSOR: Good.

  • If I were to have a matching with four edges,

  • well, there's only eight nodes, so I'd

  • have to have all eight nodes involved in the matching.

  • And that means x7 and x8 would have to be in the matching,

  • but they could only be paired with x1,

  • and so it's not possible to do that.

  • All right, so there is no matching

  • of size four in this graph.

  • Three is the best I can do.

  • Now, when you get every node in a matching,

  • then it's called a perfect matching.

  • And so in this case, it doesn't exist, but sometimes it does.

  • So a matching is perfect if it has

  • size half the number of nodes.

  • In other words, if the number of edges is v over 2,

  • then every node is in the matching.

  • All right, so that one doesn't have a perfect matching.

  • What about this graph?

  • So I got b1, b2, b3, b4, g1, g2, g3, g4,

  • and I'll put in the compatibility edges here.

  • OK, Does that graph have a perfect matching?

  • Can you pair up every boy with a girl

  • here so that everybody is compatible with their mate,

  • and you have just one spouse?

  • Can you do that?

  • AUDIENCE: Yes.

  • PROFESSOR: Yeah?

  • All right, who do I start pairing up?

  • AUDIENCE: 1, 1.

  • AUDIENCE: 1, 1.

  • AUDIENCE: b2, g3.

  • PROFESSOR: b2, g3.

  • AUDIENCE: b3, g2.

  • PROFESSOR: b3, g2

  • AUDIENCE: b4, g4.

  • PROFESSOR: There we go.

  • All right, so there is a perfect matching in this graph.

  • Very good.

  • Now, in some cases, some pairings

  • are more desirable than others, and this can

  • be represented with a weight.

  • And so you might have a weighted graph where

  • every edge has a weight on it.

  • For example, we might weight b1, g2 with 5,

  • and b1, g1 gets a 10.

  • And usually when you see weighted graphs

  • in the matching context, a lower weight

  • means it's more desirable, so that b1 and g2 would get along

  • better than b1 and g1.

  • And then the goal is to find a matching with minimum weight.

  • Now, the weight of a matching, call it M,

  • is the sum of the weights on the edges of M. Now,

  • usually when you're looking at weighted matchings,

  • you require yourself to have a perfect matching so

  • that everybody gets paired up.

  • And often in that case, you'll see all the edges present,

  • some of them with very big weights, maybe even infinity

  • if they just can't be put together.

  • Because otherwise you just say, don't match anybody together,

  • and you have weight zero.

  • So for when you look at minimum weight matchings,

  • you're looking for the perfect matching with minimum weight.

  • So we say a min-weight matching for a graph, G,

  • is a perfect matching for G with the minimum weight,

  • overall perfect matchings.

  • Let's try an example.

  • Say I've got this graph, and call this node Brad,

  • here's Billy Bob, here's Jennifer, and then Angelina.

  • And the weights are as follows-- I put a 10 here,

  • a 10 down here, a 5 here, because Brad really

  • likes Angelina and vice versa, and a 16 between Jennifer

  • and Billy Bob.

  • What is the weight of the min weight matching in that graph?

  • 20.

  • And who gets paired with who there?

  • Who does Brad get hooked up with?

  • AUDIENCE: Jan.

  • PROFESSOR: Jan, and that leaves Billy Bob with Angelina,

  • and the weight is 20.

  • So even though Brad really likes Angelina, if I go that route,

  • my weight is 21, which is not as good.

  • So the min-weight matching would be this one.

  • OK.

  • Now, it turns out that finding the maximum matching--

  • the maximum number of edges you can put together--

  • or finding the minimum weight perfect matching,

  • those are both solvable, tractable problems.

  • You don't get a million dollar prize for solving that.

  • The algorithm is run in quadratic to cubic time,

  • so not terrible, but they're not NP-complete,

  • so people know how to do it.

  • Now, they are pretty complicated,

  • so we're not going to cover them in 6042.

  • What we're going to do is look at a slightly different version

  • of the problem that actually turns out

  • to be more useful in practice, because there's

  • a very nice algorithm for it.

  • Now, in the version of the problem that we're

  • going to look at, everybody has preferences-- a preference

  • list.

  • It's not weighted, but it's a priority order

  • of who they want to get mated to, or matched up with.

  • So it would look not quite like that,

  • but it would look like this.

  • So Brad, Billy Bob, Jennifer, and Angelina-- same players,

  • but what we do is we know that Brad really likes Angelina,

  • and Angelina really likes Brad, so they are first choices,

  • at least was.

  • Number two for Brad is Jennifer, but Jennifer really likes Brad

  • first, and Billy Bob second.

  • Billy Bob likes Angelina first, then Jennifer.

  • Angelina thinks Billy Bob is number two.

  • So it's not necessarily symmetric.

  • Jennifer has Brad as the first choice.

  • Brad has Angelina as the first choice.

  • All right?

  • So it's an asymmetric situation.

  • Now, what would happen if we set up our marriages so that Brad

  • is married to Jen, and Billy Bob is married to Angelina?

  • What might happen if we made those as our pairings,

  • and we put them on a desert island, all four of them?

  • What's likely to happen there?

  • AUDIENCE: Brad and Angelina are going to cheat on each other.

  • PROFESSOR: Yeah, we're going to have a problem.

  • Because Brad and Angelina have the hots for each other here,

  • and they like each other better than their spouse.

  • So before you know it, they're going

  • to be doing their 6042 homework together late at night.

  • All right?

  • Now, when this happens, we have what's called a rogue couple.

  • Given a matching, x and y, a boy and a girl,

  • say, form a rogue couple if they prefer each other

  • over their mates in M. All right.

  • So here, if we married Brad to Jennifer,

  • and Billy Bob to Angelina, Brad and Angelina

  • form a rogue couple, because they

  • like each other better than who they were hooked up with.

  • And that's sort of a bad thing.

  • It creates instability if you were

  • to make the matchings that way.

  • In fact, we say that a matching is stable

  • if there aren't any rogue couples.

  • And now, one thing to make clear is that your preferences

  • can't change over time.

  • So it's not a situation where you get bored with your spouse,

  • and you change your mind, and then

  • you go off and create a rogue couple.

  • You're fixed in your preferences over all time,

  • no playing the field, none of that stuff here, OK?

  • So it's fixed once and for all, and your goal, of course,

  • is to create or find a perfect matching that's stable.

  • That's the goal.

  • So get everybody married up and make it stable.

  • All right, is it doable in that example?

  • If I put those four people on a desert island,

  • could I make a stable matching?

  • Yeah, who would I match Brad to?

  • AUDIENCE: Angelina.

  • PROFESSOR: All right.

  • Good, and then Billy Bob gets Jennifer.

  • Now, I'm not saying that you make everybody happy,

  • because Billy Bob and Jennifer are probably not happy there.

  • They each got their number two choice.

  • But it's stable, because Brad and Angelina

  • aren't going anywhere.

  • They are going to stay together because they

  • like each other best, and so there's no chance that Jennifer

  • is going to-- that Angelina is going to sneak off, sorry,

  • with Billy Bob, or that Brad's going

  • to sneak off with Jennifer.

  • All right?

  • So it's stable.

  • Not everybody's happy, but it's a stable set of marriages.

  • OK.

  • Any questions about what we're trying to do here now?

  • Yeah?

  • AUDIENCE: [INAUDIBLE] the edges?

  • PROFESSOR: That's a great question.

  • You'll see it referred to both ways.

  • Technically, it's a subgraph, so it has nodes and edges,

  • but you'll see me, and you'll see everybody say, oh, it's

  • a bunch of edges that don't share any nodes,

  • and you'll see them refer to it as the edges.

  • But really, underlying that, it's a subgraph, technically.

  • Any other questions about what we're

  • trying to do here to find stable, perfect matching?

  • All right.

  • Well, in this example, there was a stable, perfect matching.

  • But what about in general?

  • If I have a lot more people, and they

  • have arbitrary preferences, how many people

  • think you can always find a stable perfect matching?

  • There's one optimistic person.

  • How many people think there's some cases where you're just

  • not going to be able to do it?

  • Wow.

  • OK, it's a pessimistic view.

  • Well, in some sense you're both right.

  • If you allow boys to prefer boys and girls to prefer girls,

  • then it is not possible, and I'll give you an example.

  • You can find examples where there's always a rogue couple,

  • but if you require boys to only get matched to girls and vice

  • versa, then it is possible to always find stable marriages,

  • stable matchings.

  • And we're going to talk about an algorithm for that.

  • But before I show you the algorithm,

  • let me show you the bad case when boys can prefer boys,

  • or what's sort of a unisex scenario.

  • So here's a bad example with four people,

  • and the idea is to create a love triangle.

  • So we have Alex, who prefers Bobby Joe, Bobby Joe prefers

  • Robin, and Robin prefers Alex.

  • And their second choices go in the opposite order

  • there, all right?

  • So Alex wants to be with Bobby Joe.

  • Bobby Joe wants be with Robin.

  • Robin wants to be with Alex.

  • And then there's Mergatoid--

  • [LAUGHTER]

  • --and nobody likes Mergatoid.

  • So that's choice three for all of them here.

  • And Mergatoid's choices, preferences

  • don't really matter in this case.

  • [LAUGHTER]

  • I hope nobody is named Mergatoid in the class.

  • I would get complaints here.

  • All right, so I want to claim and prove

  • a theorem that says there is no stable matching

  • for this group of preferences.

  • So we'll state that as the theorem.

  • There does not exist a stable matching for this graph.

  • The proof is by contradiction-- assume there is one.

  • All right, so assume there exists a stable matching.

  • We're going to find a rogue couple in it.

  • So call the stable matching M. Well,

  • if there's a stable matching, Mergatoid's got

  • to get married to somebody.

  • Mergatoid will be matched with someone.

  • All right.

  • Now, here I'm going to do something

  • that you can do in your proofs, but you've

  • got to be careful when you do it.

  • I'm going to say, without loss of generality,

  • assume Mergatoid is matched to Alex, all right?

  • And I can do that.

  • So this is the abbreviation, without loss

  • of generality by symmetry.

  • And really, I should explain what I mean here in the proof.

  • Well, that love triangle is symmetric.

  • Each one has a preference for the next person

  • around the triangle.

  • In terms of graph isomorphism, with the weights on that,

  • every node looks the same.

  • So I can use symmetry.

  • So we're going to say, without loss of generality,

  • we're going to assume Mergatoid is matched to Alex.

  • And I'm implying the argument is going to be the same.

  • What I say next is the same no matter

  • who Mergatoid is matched to, because it's symmetric.

  • All right.

  • If Mergatoid is matched to Alex, do you

  • see a rogue couple up there?

  • So you have Robin matched to Bobby Joe.

  • Mergatoid is matched to Alex.

  • AUDIENCE: Alex to Bobby Joe?

  • PROFESSOR: Alex and Bobby Joe, no.

  • AUDIENCE: Between Alex and Robin.

  • PROFESSOR: Alex and Robin, yeah.

  • Alex and Bobby Joe aren't rogue, because Bobby Joe likes Robin.

  • He likes the person he or she is married to.

  • All right, so they're not going to go off with Alex,

  • but Alex and Robin are a rogue couple,

  • because Robin likes Alex the best, and Alex for sure

  • likes Robin better than Mergatoid.

  • All right?

  • So they both prefer each other to their mates,

  • and so they form a rogue couple.

  • All right, so Alex and Robin form a rogue couple,

  • and that means that M was not stable.

  • The matching was not stable.

  • And that's a contradiction, because we assumed it was.

  • So we have a contradiction, and the proof is done.

  • Any questions about that?

  • And I'm sort of implying here that you could match,

  • you could have had Mergatoid matched to Bobby Joe,

  • and then Bobby Joe and Alex would have been a rogue couple,

  • or Mergatoid matched to Robin, and then Robin and Bobby Joe

  • would be a rogue couple by the without loss of generality.

  • So it's OK to use that, but you want to be careful

  • that you're doing it OK.

  • Questions?

  • Yeah?

  • AUDIENCE: So the without loss of generality only

  • works if it's, like, perfectly symmetric?

  • PROFESSOR: Yeah.

  • Basically, the argument you're going to make

  • is just going to be the same argument done in all three

  • cases, and to save yourself some effort, you're saying,

  • don't do the three cases, do one, the other two look

  • the same.

  • So technically, you could add case one, case two, case three,

  • and they would've looked symmetric.

  • OK?

  • All right.

  • Now, this is not very surprising, as you all voted.

  • Almost all of you said it's hard to find stable matchings.

  • You might not be able to do it always.

  • And in fact, you can't in the unisex world.

  • The surprising thing is, you can always find a stable matching

  • in the world where boys could only be paired

  • with girls and vice versa.

  • Now, this statement, this result is pretty famous.

  • The problem itself is known as the stable marriage problem.

  • So let me just define it here, and then we'll

  • talk about an algorithm to find the matching.

  • So we have N boys and N girls.

  • And it's important we have the same number of each.

  • Now, actually, tomorrow, in recitation, you're

  • going to look at the scenario where there's more

  • girls than boys, or vice versa.

  • And you'll be using a similar algorithm,

  • but it'll be a different context,

  • and that's the context that comes up

  • in matching interns to hospitals, and stuff like that.

  • But for our version it's an equal number of boys and girls.

  • Each boy has his own ranked preference

  • list of all the girls.

  • So every boy sort of has his dance card

  • of the girls that he likes in order.

  • All right?

  • The orders can be different for different boys.

  • And each girl has the same thing.

  • She has her own list, ranked 1 to n of all the boys.

  • The lists are complete, and there's no ties,

  • so all the ties have to be broken here.

  • And the goal is to find a perfect matching

  • without rogue couples.

  • OK, so let's try an example.

  • Before we do the algorithm, let me just do a bigger example

  • with five boys and five girls, and we'll

  • get some feel for, this is not completely obvious how

  • to find it.

  • OK, so let's put the boys over here.

  • And here's boy 1, and his preference list

  • is going to be-- the girls will be C, B, E, A, D, and boy

  • 2 is going to have the preference list

  • A, B, E, C, D, and boy 3 is going to have D, C, B, A, E.

  • Boy 4 is going to be A, C, D, B, E, and boy 5,

  • I have to write his across like this, A, B, D, E, C.

  • And the girls also have their lists, so let's put those up.

  • So girl A likes the boys in order 3, 5, 2, 1, 4.

  • Girl B likes them in order 5, 2, 1, 4, 3.

  • Girl C has the order 4, 3, 5, 1, 2.

  • Girl D, 1, 2, 3, 4, 5, and then the last girl, E,

  • has 2, 3, 4, 1, 5.

  • All right, so say that's our matching problem,

  • and those are the preference lists.

  • Any ideas for how we might try to make an algorithm

  • to do this?

  • Any thoughts about what you'd do?

  • What are some approaches we could take?

  • Yeah?

  • AUDIENCE: Use the mating algorithm.

  • PROFESSOR: Use the greedy algorithm?

  • AUDIENCE: The mating algorithm.

  • PROFESSOR: Oh, the mating algorithm.

  • Well, yeah, the mating algorithm is going to do very well,

  • but I haven't told you what that is,

  • and it's a little complicated.

  • So I sort of want to explore the things you might know

  • from what we've done so far.

  • What approaches have we seen so far for solving problems?

  • Yeah?

  • AUDIENCE: The greedy algorithm?

  • PROFESSOR: The greedy algorithm-- yep.

  • And let's try that.

  • So let's choose the first boy, and we'll just

  • go down and give each boy the best choice available

  • and see what happens.

  • First thing to try usually is the greedy algorithm,

  • and about half the time it'll work in life,

  • and half the time it won't.

  • So let's try greedy.

  • So boy 1 is going to get his first choice, girl C. Who

  • does boy 2 get?

  • A. Who does boy 3 get?

  • D. Huh, this is going well.

  • Who does boy 4 get?

  • Boy 4, oh yeah, boy 4 has got to go all the way down to B,

  • because A, C, and D got taken, so boy 4

  • gets B. And boy 5-- who's left?

  • E, fourth choice again.

  • All right, well, maybe that's a stable matching.

  • All right, well let's see.

  • To see if it's stable, we've got to see

  • are there any rogue couples.

  • Well, is there any rogue couple involving boy 1?

  • No, boy 1 got his first choice.

  • You're not going anywhere.

  • Boy 2?

  • No-- first choice, same with boy 3.

  • These boys are quite happy.

  • They're not doing anything.

  • Boy 4 may not be so happy though,

  • because boy 4 got paired with B, so there's

  • possible rogue couples.

  • Let's see, is 4, A a rogue couple?

  • No.

  • A hates boy 4, all right?

  • No chance she's running off with boy 4.

  • AUDIENCE: 4, C?

  • PROFESSOR: 4, C?

  • All right, let's see that.

  • Where's C-- whoa, yeah.

  • C has the hots for boy 4, so she likes

  • him better than whoever her mate was,

  • and boy 4 likes C better than who he got matched with,

  • boy B. That is now a rogue couple,

  • so the greedy algorithm did not work.

  • That's too bad.

  • Well, it would be short lecture, I guess, if it worked.

  • [LAUGHTER]

  • Well, what would you do next, sort of, if you were-- I mean,

  • you could try to match 4 with C, and then try to patch things

  • up, and you could start doing that,

  • but you might create other rogue couples.

  • In fact, I don't know an approach-- yeah?

  • AUDIENCE: You could let it evolve and see

  • if it gets it more stable.

  • PROFESSOR: Say it again.

  • AUDIENCE: You could let it evolve,

  • let all the rogue couples evolve and see

  • if it ends up any more stable.

  • PROFESSOR: Yes, you could start swapping around

  • to get rid of rogue couples.

  • In doing that, you might create other rogue couples.

  • In fact, I don't know of an algorithm that

  • works like this, that works, that's known to work,

  • where you start patching things up,

  • because as you're patching things up,

  • you might make other things much worse by doing that.

  • So I don't know of an approach that way.

  • What's another approach?

  • Yeah?

  • AUDIENCE: [INAUDIBLE] pertaining to what order

  • they have each other.

  • PROFESSOR: And then do?

  • AUDIENCE: And then do the highest ordered [INAUDIBLE]

  • PROFESSOR: Min-weight matching kind of thing?

  • AUDIENCE: Maybe.

  • PROFESSOR: Maybe?

  • I don't of an approach like that that works.

  • Also, min-weight matching, that algorithm

  • is going to be more complicated than the one I'm

  • going to show you, in the end.

  • It takes more time to run.

  • And I don't even know if it works.

  • Like, I don't know if you can take these numbers

  • and make weights on the edges get a min-weight matching.

  • AUDIENCE: Could you do like a merge sort [INAUDIBLE]

  • PROFESSOR: Oh, a merge sort, so you take the minimum weight

  • edge, put that in, and recurse on that kind of thing.

  • I don't know.

  • It's possible.

  • In fact, you know what, a recursive approach

  • is a good idea.

  • I don't know of a nice recursive algorithm for this.

  • It's true that if you found a boy and a girl who

  • liked each other best, you could then

  • match them safely and recurse, because you know they're not

  • going to be in a rogue couple, because they

  • like each other best.

  • Then you could recurse.

  • But that might not exist here.

  • You might not have a boy and girl

  • that like each other best, in which case it's hard to know.

  • I don't know now if you pick the minimum weight in some sense,

  • like add the preference list or something to make a min-weight

  • and recurse on that, if that works.

  • I don't know of an approach like that.

  • But those are the kinds of things you try.

  • And as far as I know, all the simple things

  • fail for this problem.

  • But there is something that's a little more involved,

  • but does work, and that's the mating algorithm.

  • And does everybody have the handout?

  • There, it's back up there.

  • I got some copies down here if you need it, but pull that out.

  • So we're going to read this and talk

  • about what the algorithm is, and then

  • prove that it performs well.

  • So the initial condition is you have each of the N boys

  • has an ordered list of the N girls and vice versa,

  • and the ritual, we're going to view this is a mating ritual,

  • and really, the program is doing it.

  • The code is doing it.

  • But think of it as real life.

  • It takes place over several days.

  • Now, the day is broken up into three parts-- the morning,

  • the afternoon, and the evening.

  • In the morning, each girl comes out to her balcony

  • and stands on the balcony.

  • Each boy goes to the balcony of his favorite girl who

  • is still on his list that hasn't been crossed off.

  • Now, initially, every girl is on his list.

  • So he goes to his favorite girl, goes under her balcony,

  • and serenades her.

  • Now, if, over the course of the algorithm,

  • the boy has nobody left on his list, he's out of luck.

  • He just stays home and does homework.

  • All right?

  • There's no serenade anymore for him.

  • Now, in the afternoon, the girls who

  • have at least one suitor-- a boy down there serenading

  • her-- looks at all the suitors, picks her favorite,

  • and to the favorite she says, maybe I'll marry you.

  • Come back tomorrow.

  • Girls don't want to make it too easy here for the boys.

  • Now, to all the other boys who are lower priority she says,

  • I will never marry you.

  • Go away.

  • So she writes them off for good.

  • Now, that night, any boy who heard a no-- like the girl

  • said no, I'll never marry you-- crosses that girl off his list.

  • Because, you know, it's the only practical thing

  • to do at that point.

  • Now, if the boy heard the maybe I'll marry you,

  • well, he's going to go back tomorrow,

  • because that's still his favorite girl that's

  • not crossed off.

  • So he goes back and serenades her again

  • the next day in the hopes that eventually, she'll say yes.

  • Now, we keep doing this every day, OK?

  • And if we ever encounter a day where every girl has,

  • at most, one suitor, the algorithm stops, and then

  • every girl who has a suitor says, yes, I will marry you.

  • Now, if a girl doesn't have a suitor,

  • there's no one to marry her.

  • We're going to prove that doesn't happen, OK?

  • But determination condition is, you no longer

  • have a situation with two or more boys under one balcony.

  • OK?

  • All right, so let's run that algorithm on this example,

  • just so we make sure we understand it,

  • because I'm going to try to prove theorems about it.

  • So here are the serenades that are going on,

  • and here's the girls, and the days.

  • It's going to work over four days in this case.

  • And then we're also going to keep track of the boy's lists--

  • who's gotten crossed off.

  • All right, so these will be the cross outs down here.

  • Actually, maybe I'll fit it up here if I can.

  • Girls, let's see, A, B-- no, I'm going to have to space it out.

  • And the boys have their lists here,

  • and we have boys 1, 2, 3, 4, 5.

  • And here I'm going to record the cross outs.

  • All right, so let's look at day one.

  • Who is under girl A's balcony on day one?

  • AUDIENCE: 2, 4, and 5?

  • PROFESSOR: 2, 4, and 5-- each like girl A the best,

  • so she's got a lot of activity.

  • These three boys show up.

  • Anybody under girl B's balcony?

  • No, nope, no, nothing there.

  • C, does C have anybody?

  • Boy 1, yeah.

  • D?

  • 3, and E, I don't think there's any action, right?

  • Nope.

  • All right, so that's the status on day one.

  • So the action is all up here.

  • All right, so what does girl A do?

  • Who does she tell to hang around?

  • AUDIENCE: Number 5.

  • PROFESSOR: Number 5, yes.

  • She's got 5, 4, and 2, and she likes 5 the best.

  • She tells 5 to come back, and she says to these guys,

  • we're not going to marry him.

  • That means that boys 2 and 4 cross girl A off their list.

  • So they say A is no longer possible for them.

  • All right, now we go to day two, and 5 goes back

  • to girl A, and, of course, boys 1 and 3 stay there.

  • Where does boy 2 go on day two?

  • B.

  • All right, so boy 2 shows up here on day two,

  • and, let's see, boy 5 is already there,

  • and then we've got to get boy 4.

  • Where does he go now?

  • C. All right, boy 4 shows up here.

  • All right, so now the action is with girl C.

  • And what does she do?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yeah, she keeps 4, and she boots poor 1.

  • You know, led him along for a day,

  • and then he gets the boot because boy 4 showed up, right?

  • Because girl C likes 4 better than 1, so bad

  • luck for 1 there.

  • So now one goes home that night and crosses off girl C.

  • OK.

  • All right, and where does 1 go on day three?

  • All right, so he goes to-- he crossed off C-- he goes to B.

  • All right, so this is left over.

  • We have 5, 2, 4, 3-- boy 1 now goes to B, right?

  • OK.

  • And then what does girl B do?

  • Who does she keep around?

  • Keeps 2, boots poor 1.

  • All right, so 1 says, all right, I'm crossing B off my list,

  • got the message.

  • And now where does boy 1 go on day four?

  • AUDIENCE: Girl E?

  • PROFESSOR: E-- third choice.

  • OK.

  • So 1 shows up down here, and these guys keep returning.

  • Wow, so on day four there's no more fighting.

  • Every girl has at most one.

  • The termination condition is invoked,

  • and these are the marriages that take place.

  • The girls say yes.

  • All right, now is everybody--

  • [LAUGHTER]

  • --does everybody understand the algorithm we used?

  • Any questions on the algorithm?

  • Yeah?

  • AUDIENCE: [INAUDIBLE] the first case [INAUDIBLE] or

  • is that just in case the [INAUDIBLE]

  • PROFESSOR: That's in case the algorithm doesn't work.

  • Because we haven't proved it works yet,

  • and I gotta have a possibility for,

  • a boy crosses every girl off his list,

  • he gets rejected everywhere.

  • That is a possibility, so I've got

  • to say what would happen in that possibility.

  • Now, we will prove in a few minutes

  • that condition never arises, OK?

  • But I'm giving you what might happen,

  • in which so it never does, so no boy ends up

  • having to stay home and do homework here.

  • All right, so that's the algorithm.

  • Now let's see if we can see if there's any rogue couples.

  • All right, so, for example, let's look at boy 1.

  • Boy 1 got paired to his third choice, E, so 1,

  • C might be a rogue couple.

  • Is that possible?

  • No, because C got 4 here, which is her first choice.

  • She's not going with boy 1.

  • That's not working.

  • What about 1, B?

  • Could that be rogue?

  • No, B got 2, and B likes 2 better than 1, so 1, C and 1,

  • B are not rogue.

  • And 1 got E, so there's no other possibility for boy 1,

  • so boy 1 is not in a rogue couple.

  • What about boy 2?

  • 2 got B, and that's 2's second choice.

  • What about 2, A?

  • Let's see, A got 5, and A likes 5 better than 2,

  • so that's not rogue.

  • Boy 2 is not rogue.

  • All right, boy 3-- boy 3 got his first choice, right?

  • 3 got D, so 3 is not going anywhere.

  • Boy 4 got-- who did boy 4 get-- got C. 4 got his second choice.

  • What about 4, A?

  • A hates 4.

  • A is not gonna get caught dead with 4, so that's not rogue.

  • So 4 is OK.

  • 4 is not in a rogue couple.

  • And finally, boy 5-- boy 5 is paired with A,

  • and that is boy 5's first choice,

  • so he's not wandering off here.

  • All right?

  • So in fact, we've just argued that this

  • is a stable set of marriages, a stable matching in this case.

  • Any questions now about what we're trying to do?

  • OK, because we're going to try to show now

  • it always produces a stable matching.

  • And to do that, we need to do a few things.

  • All right, so what are the things

  • we need to show to prove everything

  • is going to be good here?

  • What are some facts we to prove?

  • Yeah?

  • AUDIENCE: Don't we need to show that the algorithm, it does

  • come to an end?

  • PROFESSOR: Yes.

  • [LAUGHTER]

  • The algorithm terminates, so we need

  • to show that the marriage algorithm, TMA, terminates.

  • Otherwise, the boys are serenading forever,

  • and that's not too good.

  • All right, what else do we want to show?

  • Yeah?

  • AUDIENCE: If it does terminate, then one is left empty.

  • PROFESSOR: Then?

  • AUDIENCE: Everyone gets someone.

  • PROFESSOR: Everyone gets married.

  • Yep.

  • Stability is easy if nobody gets married.

  • All right, what else do we want to show here?

  • Yeah?

  • AUDIENCE: There are no rogue couples.

  • PROFESSOR: No rogue couples.

  • All right, what else might we like to show?

  • These are the three main ones.

  • There's a couple of other things you might like.

  • Yeah?

  • AUDIENCE: It runs quickly.

  • PROFESSOR: It runs quickly.

  • Well, you can only serenade for so long.

  • In fact, it does run quickly, and that's

  • why it's useful in practice.

  • Anything else you might want to show about this algorithm?

  • AUDIENCE: How many people you crossed out?

  • PROFESSOR: How many people you crossed out-- yeah,

  • you could, and that'll tie into how long it takes.

  • Yeah.

  • Anything else?

  • Yeah?

  • AUDIENCE: The average likeness between the couples.

  • PROFESSOR: The average likeness, oh.

  • So yeah, we haven't had a notion here

  • of how happy people are at the end,

  • because we don't have a weighting on the edges,

  • but you might want to think about that.

  • That's actually a good point.

  • Yeah?

  • AUDIENCE: If they [INAUDIBLE] if it

  • matters who serenades [INAUDIBLE] girls or boys

  • to get her.

  • PROFESSOR: That's a great point.

  • Is this algorithm good for girls, or good for boys?

  • Yeah, that's a good point-- fairness.

  • All right, so we'll take a look at fairness also.

  • Is it better to be a serenader, or be on the balcony making

  • your choices?

  • OK, so this is what we've got to do.

  • So let's start by showing TMA terminates,

  • and that it terminates pretty quickly.

  • Now, in fact, I'm going to prove a fairly crude bound

  • on the time, but it actually does fairly well.

  • So our first theorem is going to be

  • that TMA terminates in, at most, N squared plus 1 days.

  • N is the number of boys and girls.

  • The proof is going to be by contradiction.

  • This is probably the only day where

  • we'll do a bunch of proofs and none of them use induction.

  • They're pretty much all by contradiction.

  • Suppose TMA does not terminate in N squared plus 1 days,

  • because we're going to show that leads to a contradiction.

  • We need to show that some kind of progress

  • is made each day to show that it terminates.

  • Yeah?

  • AUDIENCE: [INAUDIBLE] the number of causes that have

  • [INAUDIBLE], so if it has not terminated,

  • then at least one girl is seeing a group of at least one guy--

  • PROFESSOR: Yes.

  • AUDIENCE: --so the number of crosses cannot be defined

  • [INAUDIBLE]

  • PROFESSOR: Very good.

  • OK, let's state that as a claim, and you've

  • given the proof of the claim, which is great.

  • If we don't terminate, on a day, that

  • must be because a girl had two boys there, or more, therefore

  • she rejected some, at least one.

  • And that night, the rejected boy crosses a girl off his list.

  • So if we don't terminate, then at least one boy

  • crosses at least one girl off his list.

  • And so we're going to measure progress by the cross outs.

  • So every day we don't terminate, a boy

  • crossed a girl off his list, so if we didn't terminate

  • at N squared plus 1 days, we must

  • have crossed off N squared plus 1 girls across all the lists.

  • Well, is that possible?

  • To have done N squared plus 1 cross outs?

  • What do you think?

  • How many names are on each list?

  • AUDIENCE: N.

  • PROFESSOR: N, and how many lists are there?

  • AUDIENCE: N.

  • PROFESSOR: N, so there's N lists with N names implies

  • there's, at most, N squared cross outs ever.

  • But we just said we had N squared plus 1 cross outs.

  • But we have also N squared plus 1 cross outs,

  • and that's a contradiction.

  • All right?

  • So we're done.

  • It has to terminate within N squared plus 1 days.

  • Any questions?

  • This is a very common proof technique in computer science.

  • You're analyzing some system, and every step

  • or every day or every time period,

  • you want to argue progress got made,

  • and then after you've made enough progress you

  • have to be done, and therefore, the algorithm is completed.

  • All right, so we know that TMA terminates.

  • Now, we've still got to get everyone married

  • and have them all be happy, or at least stable.

  • Now, to do this, we're going to use an invariant.

  • Yeah?

  • AUDIENCE: If a girl has some guy in front

  • of [INAUDIBLE] on something, then she

  • will always have [INAUDIBLE]

  • PROFESSOR: That is true.

  • Something even stronger is true.

  • AUDIENCE: All the [INAUDIBLE]

  • PROFESSOR: Have a--

  • PROFESSOR: If she has a preference for someone who

  • [INAUDIBLE] only better guys.

  • PROFESSOR: Yeah, that's a great invariant.

  • As the girl is sitting there on her balcony,

  • things only get better, because she always

  • keeps the best one around that's there.

  • And when new ones come in, they've

  • got to be better than the last one for her to keep them.

  • So as she rejects boys, she only does that because she's

  • got better ones there, and whoever she says maybe to

  • always comes back.

  • So an invariant of this algorithm

  • is that when a girl has a suitor, going forward,

  • she only has suitors she likes at least as well.

  • And if she ever rejects a boy, then she's

  • got somebody better there forever.

  • All right, so let's state that as an invariant and prove that.

  • OK, so we're going to let P be our invariant here.

  • P is the statement that if a girl, G, ever rejected a boy,

  • B, then the girl, G, has a suitor,

  • or if the algorithm is terminated,

  • a husband who she prefers to B. All right,

  • that's going to be our invariant,

  • and now we've got to prove it's an invariant.

  • So we'll do that with a lemma called lemma one.

  • P is an invariant for TMA.

  • All right, let's prove that.

  • Now, what's the first thing you've

  • got to prove when you're proving something as an invariant?

  • There's two things you've got to do to establish a variant.

  • What's the first one?

  • AUDIENCE: Base case?

  • PROFESSOR: Base case.

  • Show it holds true at the beginning.

  • Proof is going to be by induction,

  • and so we've got to show that P holds true at the beginning.

  • But what are we going to induct on here?

  • What's the parameter we're inducting on?

  • AUDIENCE: Time?

  • PROFESSOR: Time, the number of days.

  • Good.

  • The base case is day zero, the beginning.

  • All right.

  • Well, is that statement true on day zero?

  • Yeah, it's true for sort of a weird reason.

  • Nobody's been rejected yet, so it's

  • what's called vacuously true.

  • Because no girl, G, has rejected any boy, so it's true.

  • So no one is rejected yet.

  • So it's vacuously true.

  • All right, next we have the induction step.

  • So we'll assume P holds at the end of day d,

  • and we need to argue that it holds now

  • at the end of day d plus 1.

  • So say it's true up to now, up to day d.

  • Why Is it true at the end of the next day?

  • Well, there's two cases to look at here, depending

  • on when G rejected B. So if she rejects B on this day, day

  • d plus 1, well, why would she reject B on day d plus 1?

  • There's only one scenario.

  • Yeah?

  • AUDIENCE: There's a better boy.

  • PROFESSOR: There's a better boy.

  • And so she says maybe to him, and he becomes her suitor.

  • So, in fact, P is true.

  • So then there was someone better,

  • and that implies P is true on day d plus 1.

  • Case two is very similar.

  • It's a very simple proof.

  • G rejected B before day d, before d plus 1.

  • Well, now we use the fact that P was true at the end of day

  • d, which means P now implies that G had at least as good

  • a suitor.

  • Actually, it's better, because it was rejected.

  • A better suitor on day d, that's what the hypothesis says,

  • the invariant says.

  • And now we just have to look at what happens on d plus 1.

  • Well, either she has the same suitor on d plus 1,

  • or somebody better came along.

  • And so we're going to be done.

  • She has the same or better suitor on day d plus 1,

  • and that implies P is true on d plus 1, and we're done.

  • All right, so I went through this proof.

  • It was sort of obvious, but this is

  • the careful way you'd write it down

  • to show that the invariant holds.

  • Any questions about the invariant dilemma?

  • All right, so things only get better for the girls.

  • If she ever rejected somebody, she's got somebody better.

  • So now we can prove the main result,

  • that everyone is married.

  • And again the proof will be by contradiction.

  • So we assume not everyone was married.

  • Assume, for the purpose of contradiction,

  • that some boy, B, we'll call him, is not married.

  • Because if everyone is not married then some boy is not.

  • If not everyone is married then some boy is not married.

  • So when it terminates, B is not married.

  • Well, what do you know about B if he was not

  • married at the end?

  • PROFESSOR: He was rejected by everyone.

  • PROFESSOR: He was rejected by everyone,

  • because if at the end, he's still under a balcony,

  • if he still had somebody on his list, he'd be there.

  • And he'd be getting married, because it's the end.

  • So this means that if B is not married,

  • he's rejected by everybody.

  • Yeah?

  • AUDIENCE: [INAUDIBLE] list?

  • PROFESSOR: Oh, he's on everybody's list.

  • AUDIENCE: Yeah.

  • PROFESSOR: Everybody has everybody

  • of the opposite sex on their list,

  • but he had to cross everybody off.

  • That's true.

  • Yeah?

  • AUDIENCE: That would mean that everyone had somebody

  • better on the same day that [INAUDIBLE] telling him she

  • didn't want to be with him.

  • PROFESSOR: That's true.

  • That means that B crossed every girl off.

  • B is rejected by every girl, which

  • means every girl has somebody better

  • than B, which is not possible, because that would

  • mean every girl was married.

  • And therefore, the equal number of boys and girls, that means B

  • would have been married.

  • Good.

  • All right, let's write that down.

  • This means that B was rejected by every girl.

  • OK, that means that every girl, by lemma 1,

  • has a better suitor, and that's where we use lemma 1.

  • And that means that every girl is married,

  • and that means that every boy is married, including B,

  • and that's a contradiction, because we

  • said B wasn't married.

  • OK?

  • Everybody buy that proof?

  • Any questions about that?

  • Yeah, proof by contradiction is a pretty powerful technique.

  • Once you assume something is not going to be true,

  • it gives you a lot of power to find a contradiction.

  • All right, so now we know that the algorithm ends,

  • and everybody gets married.

  • All's we got to do is show that there's no rogue couples,

  • so let's do that.

  • TMA produces a stable matching.

  • Now, how do you suppose we're going to prove this?

  • What's going to be the approach to prove this?

  • Yeah?

  • AUDIENCE: Assume that there's a rogue couple.

  • PROFESSOR: Assume that there's a rogue couple,

  • that namely this is not true, so there must be rogue couple.

  • So let Bob and Gail be any pair that are not married.

  • I need to prove the Bob and Gail are not rogue,

  • and then we'll be fine.

  • Because if it says everybody who is not married is not rogue,

  • then we know we have a stable matching.

  • Now, there's a couple of cases here.

  • Bob and Gail weren't married, so there's two ways

  • that could have happened.

  • What's one of them?

  • What's one reason they might not be

  • married, something that happened that made that impossible?

  • Yeah?

  • AUDIENCE: Gail rejected Bob.

  • PROFESSOR: Good.

  • Case one-- Gail rejected Bob.

  • Well, what do we know in that case, if Gail rejected Bob?

  • AUDIENCE: Gail had better suitors.

  • PROFESSOR: Gail has another suitor that she likes better.

  • And, in fact, what do we know about who Gail married?

  • AUDIENCE: Better than Bob.

  • PROFESSOR: Better than Bob-- things only

  • get better for the girls.

  • That's the lemma one.

  • All right, so this means that Gail marries someone that she

  • thinks is better than Bob.

  • And that's by lemma one.

  • Well, can Gail and Bob be a rogue couple here?

  • No, because Gail likes her spouse better than Bob,

  • so she's not going to be in a rogue affair with Bob.

  • So that means that Gail and Bob are not rogue.

  • All right, case two is Gail did not reject Bob.

  • She never did.

  • That's the other case, she didn't reject Bob.

  • Could Bob have ever serenaded Gail in this case?

  • No, because if he did and he was never rejected,

  • they would have ended up married.

  • All right?

  • So that means the Bob never serenaded Gail.

  • What does that mean about how Bob feels about Gail?

  • Yeah?

  • AUDIENCE: Bob must have never serenaded for Gail.

  • PROFESSOR: Yeah, Bob never got far enough down

  • on his list to serenade Gail.

  • He got married before he got down there.

  • All right, so that means that Gail is lower

  • on Bob's list than Bob's wife.

  • And that means that they're not rogue,

  • because Bob likes his wife better than Gail.

  • All right, so in each case, the cases clearly cover everything.

  • Gail rejected Bob, or she didn't.

  • Either way, Gail and Bob are not rogue.

  • So that means there's no rogue couples,

  • and that implies M is stable, TMA is stable.

  • OK?

  • All right, so TMA terminates, everyone

  • gets married, no rogue couples, nice outlook.

  • So we're done.

  • We actually proved it works.

  • One issue left to think about here.

  • Any questions on that before we launch off into the last issue?

  • AUDIENCE: Is it unique?

  • PROFESSOR: Is it unique?

  • TMA gives you a unique answer because it's an algorithm.

  • It's deterministic.

  • But there may be other stable matchings.

  • OK, so there's not just one stable matching, necessarily.

  • You could make examples with multiple stable matchings.

  • That's a great question.

  • Any other questions?

  • All right.

  • Oh, yeah?

  • AUDIENCE: Is there generally any other way

  • to assess optimality besides the fact that it's stable, or--

  • PROFESSOR: Yeah, you can make up lots of them.

  • You could put weights on the edges

  • and get a min-weight matching.

  • You could try to get the perfect matching

  • with the least unfavorable marriage kind of thing.

  • There's a lot of criteria you can make.

  • This one turns out to be useful in practice in a variety ways

  • that we'll talk about, and also have

  • a nice, fast, simple algorithm that can actually

  • run in a distributed environment, which

  • is really nice.

  • So it is probably the most practical approach

  • to matching out there.

  • Any other questions?

  • OK, the last issue-- I don't know

  • if we still have the issues up there-- is fairness.

  • So who thinks the TMA is favorable to the boys?

  • Just a couple.

  • Who thinks it's favorable to the girls?

  • More, that's the common response.

  • Who thinks you can't even define it one way or the other?

  • It's unclear and hopeless to decide.

  • OK, well, it seems like maybe the girls, because they

  • get the best of their suitors.

  • They sit back, and they just take the best

  • as they come along.

  • On the other hand, the boys do try

  • to go out and get their first choice.

  • The girls have to wait.

  • And Mr. Right may never come along.

  • The boys are out there.

  • I'm going to my first choice, and they get denied, OK, they

  • just move right on to the next choice.

  • So this is actually one of these questions

  • of study in sociology.

  • In the animal species, which is better, proposers or acceptors?

  • What's the more powerful result?

  • Who has the better power in courtship?

  • It turns out that we can answer this question in a very

  • clear way, and prove it here.

  • And the answer is the boys have all the power here.

  • This is very favorable to the boys, and we'll see why.

  • Now, to prove that, we need some definitions.

  • OK, so I've got to define a couple things here

  • to be able to prove this.

  • The first is, for any collection of preference lists,

  • we're going to let S be the set of all stable matchings.

  • Now, we know that S is not empty, right?

  • How do we know that S is not empty here?

  • AUDIENCE: [INAUDIBLE] stable.

  • PROFESSOR: Yeah, because TMA produces a stable matching,

  • so we know that S is not empty.

  • There's at least one.

  • And, in fact, there could be many.

  • Now, for each person, P, we define the realm of possibility

  • for P to be the set of mates that you might

  • have in a stable matching.

  • So it's a set queue of people for which

  • there exists a matching that's stable such that you're

  • mated to that person.

  • So I've written this in some-- that's

  • sort of complicated mathematics, but somebody

  • is in your realm of possibility if there is a stable matching

  • where you married them.

  • If it exists, there's a stable matching

  • where you could marry them.

  • And this is vaguely like you see sometimes, you're

  • going out with somebody and your parents say, oh, they're not

  • in your league, or something.

  • This is a mathematical formulation

  • for that, that there is some way in the world to marry everybody

  • up, so it's all stable, and you could

  • be married to that person.

  • OK?

  • Let's do an example with four people.

  • All right, so for example, say we have Brad, Jen, and Angelina

  • again.

  • And of course Brad likes Angelina and vice versa.

  • Gen likes Brad, and then there's Billy Bob here.

  • All right.

  • Now, Brad is not realistic for Jen.

  • Brad Is not in Jen's realm of possibility,

  • because in any perfect matching where Brad marries Jen,

  • you're going to have a rogue couple.

  • Brad's going to go for Angelina.

  • So Brad is not within Jen's realm of possibility.

  • And similarly, vice versa-- Jen is not

  • in Brad's realm of possibility, because there's no stable

  • matching where they're married, in this scenario,

  • for this problem.

  • All right?

  • Now that we have that notion, we can

  • define who your optimal mate is, and who your pessimal mate is,

  • the worst case.

  • All right, so we define a person's optimal mate

  • is his or her favorite in the realm of possibility.

  • So your optimal mate is not your favorite overall,

  • it's just your favorite among those who are possible,

  • that there is a stable matching that you could

  • be matched to that person.

  • And similarly, you have a person's pessimal mate

  • is your least favorite from the realm of possibility.

  • OK?

  • Because you have all the people you

  • might be married to in stable matchings.

  • Your favorite is the optimal, your least favorite

  • is the pessimal.

  • Any questions about that?

  • Does that make sense?

  • And you don't count the ones you can't

  • be married to, because if you were, it would be unstable.

  • OK, now we can state the blockbuster theorems here.

  • Theorem four says TMA marries every boy

  • with his optimal mate.

  • Theorem five, probably can guess,

  • TMA marries every girl with her pessimal mate.

  • All right, so it is optimal for every boy.

  • They get the best possible they could have in stable marriages,

  • and every girl gets the worst possible mate.

  • Now, you wouldn't necessarily think that when you first

  • look at that algorithm, where the boys are working down

  • their list, and the girls are just getting better and better.

  • But that's the case.

  • Now, let's see, I'm not going to have time to prove them both,

  • but I will prove theorem five.

  • And I'm going to assume theorem four is

  • true to prove theorem five.

  • Maybe we'll do theorem four tomorrow in recitation.

  • I don't know.

  • So let's do theorem five so you can see how this works.

  • The proof is by contradiction, so we'll

  • assume theorem four is true, and we'll

  • prove the proof of theorem five by contradiction.

  • So assume there was a stable matching where a girl got

  • worse off than in TMA.

  • All right?

  • Assume three and five is not true,

  • then there's some girl in some stable matching.

  • So suppose, for the purpose of contradiction,

  • that there exists a stable matching,

  • we'll call it M, where some girl,

  • and she fares worse than in TMA.

  • I want to show that results in a contradiction.

  • And the proof is pretty simple, just a simple picture.

  • So G did worse in M than in TMA.

  • All right, so let's let B prime be the mate of G in M.

  • So girl G did worse in M than in TMA, so here's her mate in M,

  • and let B be her mate in TMA.

  • Who does she like better, B or B prime?

  • B, because we're saying that she did worse in M than she did

  • in TMA, so she likes B better.

  • And let G prime be the mate of B in M. Now, who does

  • B like better, G or G prime?

  • [INTERPOSING VOICES]

  • PROFESSOR: Not G prime.

  • B got married to G prime in M, but what does theorem four say?

  • Yeah, G. Theorem four says that across all

  • the stable matchings, B gets his favorite mate in TMA.

  • So G prime is in the realm of possibility, and so is G,

  • and we know from theorem four B likes G

  • best, better than G prime, because of theorem four.

  • So what happened here?

  • There's a rogue couple in M. B and G are a rogue couple in M.

  • And that means that M is not stable,

  • and that is a contradiction.

  • All right?

  • And so theorem five is true, because we

  • assumed we had a stable matching M where there's

  • a girl who did worse off.

  • So it really pays off to be aggressive party in courtship.

  • Everybody lives happily ever after,

  • especially the boys in TMA.

  • Now, TMA arises in lots of applications.

  • The most famous is the matching program

  • that's used to match MDs to residency programs.

  • So how many pre-meds are here?

  • There's one, at least.

  • So you're going to go to medical school some day,

  • and at the end of medical school,

  • there's a big day called match day,

  • where you get assigned to a hospital for your internship,

  • and the way that works is using this algorithm.

  • And who do you suppose is the boy in this algorithm?

  • AUDIENCE: The hospital.

  • PROFESSOR: The hospital.

  • And they want it so that when they make the matchings,

  • that there's not some pair of an intern or doctor in a hospital

  • where they'd like each other better than what they got.

  • This is used in online dating for the obvious reasons,

  • and we actually use this algorithm lot at Akamai

  • to load balance traffic on the web.

  • And here you have the boys are web servers,

  • and the girls are requests for service.

  • And the goals are to balance performance,

  • getting a server that's nearby you that's fast.

  • And on the other side, our cost, and by who

  • we make a boy and a girl, we can trade off cost for performance

  • in a very nice, distributed way.

  • All right, so that's it for today.

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