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Lecture 02 : The Stable Matching Algorithm — Transcript

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0:01[Music]
0:22[Music]
0:27welcome to lecture two of artificial
0:29intelligence for economics in lecture
0:31one we looked at two examples of network
0:34data of interpreting network data in
0:37today's lecture we'll look at something
0:39completely different we look at what's
0:41called the stable matching
0:44algorithm now what is
0:47that it's basically a matching problem
0:51uh as the name suggests but what exactly
0:54do we mean by
0:55that uh let's say we have two
0:57heterogeneous populations X and Y
1:01okay uh it could be a set of boys and a
1:04set of
1:05girls every element in X or every member
1:08of set
1:09X has a preference ordering over the
1:12elements of set
1:15y similarly every element of set y has a
1:20preference ordering over the elements of
1:22set
1:23X if that's the case who should be
1:26matched with whom that's the kind of
1:29question
1:30which we want to answer in this
1:32lecture where do we see situations like
1:36this well in the in the marriage or
1:38dating Market let's say there are a
1:40bunch of men and a bunch of women Every
1:43Man Has a preference ordering over the
1:46set of women and every woman has a
1:48preference ordering over the set of
1:50men labor market let's say there are ex
1:53CEOs and X firms or Y
1:55firms and the CEOs have a preference
1:59ordering over the set of firms and every
2:01firm has a preference ordering over the
2:03set of individuals or
2:06CEOs credits and Banks right so there
2:10are firms and Banks the firms have a a
2:13preference ordering over which bank they
2:15want to borrow
2:17from and uh the banks also have a
2:19preference ordering over the set of
2:22firms similarly have we have buyers and
2:25sellers so on and so forth so these are
2:28typical examples of where we uh
2:32encounter a matching the need for
2:38matching there could be many to One
2:40problems too for example campus
2:43placements there are students and then
2:45there are firms who visit the
2:48campus the the students uh definitely
2:51have a preference ordering over the set
2:54of firms who have come to
2:57recruit the firms also have a preference
3:00ordering over the set of students based
3:02on their GPA or or other
3:06credentials
3:08great so let's think of a particular
3:13context let's think of marriage or the
3:15dating Market which is the first example
3:17I
3:19cited so we are going to make a
3:20simplifying assumption we'll make an
3:22assumption of monogamy that is one to
3:24one matching one boy for one girl one
3:27girl for one
3:28boy we also assume equal sized
3:31population that is the set of boys and
3:34the set of girls are they have same
3:38cardinality if that's the situation the
3:41question is how to optimally
3:44match right but when am I when I'm using
3:47the word optimal what exactly do I mean
3:50by that what do I mean by Optimal let's
3:54look at an
3:58example consider
4:01a set of two boys and two girls the two
4:04boys are named Rahul and Aman and the
4:07two girls are let's say Angeli and
4:10Tina now every boy remember has a
4:13preference ordering over the set of
4:15girls so Rahul prefers uh Tina the most
4:20and then anjeli Amon also prefers Tina
4:24the most and then
4:25aneli Tina prefers Rahul and then Amon
4:28aneli prefers Rahul and then then number
4:30okay and this can be uh represented by
4:33this uh easy graph or network whatever
4:37you might call
4:39it now the question of this if this is
4:42the preference orders of the different
4:45individuals how to optimally
4:48match what are the possible
4:51matchings well these are my two possible
4:54matchings Rahul matched with aneli Aman
4:57matched with Tina or Rahul matched with
5:00Tina and Ammon matched with aneli
5:03correct
5:04F let's look at matching one the first
5:09matching Rahul analii and am amantina
5:13let's look at this
5:15matching is there any problem with this
5:17matching let's understand let's look at
5:20the preference
5:23relationships preference
5:25orderings Rahul and aneli and Amon and
5:28Tina these are my two pairs
5:31who does Rahul prefer the most Rahul
5:34prefers Tina the
5:35most who does Tina prefer the most Tina
5:38prefers Rahul the most so in matching
5:42one Rahul is paired with anjeli but
5:46Rahul likes Tina more Tina is paired
5:49with Ammon and Tina likes Rahul more
5:51than
5:52Ammon so in this case both Rahul and
5:58Tina uh uh prefer each other over their
6:03current mates or current
6:06Partners which means they will break out
6:09of their current
6:10Partnerships right Rahul is paired with
6:13anjeli Rahul has all the incentive to
6:16break out from that Partnership if Tina
6:19says
6:20yes because why because Rahul prefers
6:22Tina more than
6:23anjeli Tina on the other hand uh has an
6:27incentive to break out of her her
6:30partnership with Amon because Tina
6:32prefers Rahul more than Amon so Tina
6:36would be more than eager to break out if
6:37Rahul says yes now both Rahul and Tina
6:40are eager to break out so both of them
6:42will say yes and they will break out of
6:44their current Partnerships so matching
6:47one is
6:50unstable
6:53okay we
6:56say that uh given a matching M two
7:01individuals X and Y form a rogue couple
7:05if they prefer each other over their
7:08mates so in this case Rahul and Tina in
7:12matching one Rahul and Tina former Rog
7:18couple okay matching two by the way has
7:21no Rog
7:25couples great now that we know what a
7:28Rog couple means here's the definition
7:31so what is a stable
7:34matching a stable a perfect matching is
7:36where all individuals are paired
7:39okay a stable matching is a perfect
7:42matching such that there are no row
7:46couples so matching one is not a stable
7:50matching matching two on the other hand
7:53is a stable
7:56matching great let's move on
8:01let's look at one more
8:03example let's take a let's take a movie
8:06example let's say we have these four
8:08movie stars Akshay Salman amitab and
8:12John and uh let's say there are two
8:16movies which are being
8:17made and in each movie there will be two
8:20stars okay now who will be paired with
8:25whom now what is their preference
8:27relation preference ordering
8:30well akshai prefers amitab the most then
8:32Salman then John as you can see from
8:35here right Akshay prefers amitab the
8:38most this is one I'm sorry
8:44uh so AKA prefers amitab the most then
8:48Salman and then
8:50John right similarly uh Salman prefers
8:57Akay then amitab then John am prefers
9:00Salman then AK then John well John's
9:03preferences are inconsequential because
9:04he's everybody's last
9:07choice I mean no no offenses against uh
9:10John but then
9:11yeah uh this is just
9:15hypothetical okay let's move
9:19on the theorem is there does not exist a
9:22stable
9:23match if this is the preference
9:26ordering and if we want to pair if we
9:28want to form two
9:30pairs there we can't we can't do it we
9:33can't form a stable
9:35match okay let's
9:40see assume that there exists a stable
9:42match n so we are going to prove by
9:44contradiction
9:46here assume that there exists a stable
9:50match so without loss of generality JN
9:54will be matched with somebody in that
9:56stable matching let us say John has been
9:59matched with AA without loss of
10:02generality wlog is without loss of
10:06generality now if Jon is matched with
10:11AKA then by default Salman is matched
10:14with
10:16amitab
10:21great but then look at the look at the
10:23preference relations preference
10:26orderings so Salman is matched with
10:28amitab AKA is is matched with John let's
10:30go back to our preference
10:37ordering uh Salman is matched with
10:39amitab and aksha is matched with John
10:42but aksha prefers amitab
10:46more uh Salman
10:50prefers uh sorry AA is matched with John
10:54so AKA prefers John the least so AKA
10:57will definitely want to move out
11:00right Salman on the other hand prefers
11:03aka the
11:05most okay so aksha and Salman will form
11:10a rogue
11:13couple
11:15right AKA has been matched with JN let's
11:18say without loss of generality then aka
11:20prefers John the least so aksha wants to
11:22break
11:23out and Salman prefers a aka the most so
11:28Salman and Akay will form a Rog couple
11:31Salman also would want to break out
11:34because he prefers Akshay more than
11:37amitab
11:38great so this is not there can't be a
11:41stable match so m is not
11:44stable and we can prove
11:46this uh for if I if I match John with
11:51this is an exercise for all of
11:52you if you match John with Salman and
11:56then see if you can find a rogue couple
11:58you will be able to find one
12:00no matter whom you match John with there
12:03will be a Rog
12:05couple great so the this was my
12:07preference
12:09orderings I can't find a stable
12:12match well this just gives you an
12:15inkling towards a more General result
12:17which I'm going to talk about
12:20now this is the more important theorem
12:22this theorem states
12:25that a stable match will necessarily
12:28exist
12:29if the preference list can be
12:31represented by a bipartite graph what is
12:33a bipartite
12:35graph that is if we have two mutually
12:38exclusive sets
12:41and any member of set one has a
12:45preference ordering over the individuals
12:47of set two and any individual of set two
12:51has a preference ordering over the
12:52individuals of set one in such a
12:55setting a stable match necessarily
12:58exists
13:01okay if this case which we just talked
13:05about it was not a bipartite scenario
13:09right we can't find two distinct two
13:12mutually exclusive sets such that every
13:15individual in that set has a preference
13:17ordering over others it's not the case
13:19here this is not a bipartite
13:23graph okay so now this is theorem two
13:29now we have seen that if it if it is not
13:32bipartite we have found an example where
13:35a stable match does not exist but does
13:39it tell us for sure that if the if the
13:43if there is a bipartite graph
13:44representing the preference
13:47orderings uh then we will necessarily
13:51have a stable
13:52match the answer is yes and we are going
13:54to prove
13:55that okay so in the next part of the
13:59lecture what we'll do is the following
14:01we'll try to prove that in such a
14:04scenario a stable match necessarily
14:07exists and we'll also propose an
14:10algorithm to find that stable
14:14match okay but we'll go the other way
14:17around we'll first propose the
14:20algorithm and then claim that the
14:23algorithm works we'll first propose an
14:26algorithm to find the stable match and
14:29then prove the existence of the stable
14:31match by proving that the algorithm
14:33necessarily Works under all
14:36scenarios okay great so first uh I'll
14:41try to propose an algorithm for
14:43finding a stable
14:46matching but while proposing the
14:48algorithm I will take help of an
14:51example so let's say we have five girls
14:54and five boys the girls are named a b c
14:57d e and the boys are named 1 2 3 4
15:035 these are my preference
15:07orderings every boy has a preference
15:09ordering over the girls so this is boy
15:12one's preference ordering let's say so
15:14boy one prefers girl C the most and then
15:19B and then e and then a and then D okay
15:22similarly every boy has a preference
15:23ordering similarly girl every girl also
15:26has a preference ordering over the set
15:29of
15:30boys girl a for example
15:33prefers boy three the most and boy four
15:37the
15:38least
15:41okay fine let's move on so if this is
15:44the
15:45situation can we find a stable matching
15:49that is a matching where there will be
15:52no Rog
15:55couples so
15:57first which is what usually uh we try to
16:01do we'll try to propose a greedy
16:03algorithm a greedy algorithm is the most
16:05intuitively obvious algorithm what is a
16:08greedy algorithm a greedy algorithm is
16:10where we optimize stepwise and we don't
16:13go
16:14back okay so we'll propose a greedy
16:17algorithm and we'll we'll try to guess
16:19or we'll try to see if this greedy
16:21algorithm works or
16:23not great let's begin start with boy one
16:27so this is the algorithm we start with
16:29boy one and allocate the best possible
16:33girl that is what is best possible girl
16:36the highest in his preference
16:38list allocate him that
16:42girl
16:45next look at boy two and match him with
16:48the best available girl again what what
16:51do what do I mean by best according to
16:54his preference list remember these are
16:56the preference lists so what's going to
16:59happen so who does boy one
17:04prefer by the way and we are going to
17:06carry on like this so let's see what the
17:09what outcome the greedy algorithm gives
17:12us so who does boy one prefer the most C
17:16so I'm going to match one with C that's
17:19what I do one likes C the most one with
17:22C then I'll come to boy two who does boy
17:27two like the most a a is a available
17:30that is is a already matched no a is
17:33available so I'm going to
17:35match two with
17:39a okay so C and A have been taken now I
17:43come to three boy 3 who does boy 3 whom
17:47does boy 3 like the most D girl
17:51D is girl D
17:53available yes only C and A have been
17:56already matched girl D is available so
17:59I'm going to match three with
18:01d great so c a d have been
18:04taken now I come to boy
18:07four who does boy four like the most a
18:10but a has already been matched with two
18:13so I can't do
18:14that next C well C has already been
18:17matched with one so I can't do that then
18:20comes D well D has already been matched
18:22with three I can't do that either so
18:25four will be matched with the best
18:27available girl which is
18:31B right and coming to five five will be
18:35matched with the only girl who is left
18:37which is
18:40e okay so that's it that's the match
18:44which we
18:45get uh by using the greedy
18:48algorithm but now the question is is
18:51this matching which we have got is this
18:53a stable
18:55match how do we inspect that we try to
18:58look at these couples and see if any of
19:01these
19:03couples form or is a rogue
19:08couple okay so let's let's
19:12inspect and it turns
19:15out that boy four and girl C form a
19:19rogue
19:20couple let's understand
19:23why let's look at boy four and girl C
19:27who four matched with four is matched
19:30with b and one is matched with
19:34C
19:36fine
19:38now who does four prefer the
19:42most four prefers four is now hitched
19:46with
19:48B right but four prefers C more than
19:53b four prefers C more than
19:56b c right now
20:00is matched with
20:04one but C prefers four the most see look
20:09at C's preference ordering C prefers
20:12four the
20:14most and four prefers C more than his
20:18own partner which is
20:21B so c will definitely want to break out
20:24because C prefers four the most and four
20:26will also break out because four refers
20:28C more than the current partner which
20:31four has which is
20:32B so which means boy four and girl c
20:36will form a Rog couple so we see that
20:39the greedy algorithm gives us a a
20:43matching which is not
20:46stable okay so the greedy algorithm has
20:49failed so what do we do we naturally
20:53we'll have to propose an alternative
20:56algorithm so here we are proposing
20:59are stable matching
21:02algorithm please understand this
21:04algorithm carefully you can pause the
21:06video and read the slide or listen to
21:08this uh once more so this is how the
21:12algorithm
21:14goes every day a boy will go and stand
21:17in front of the balcony of the
21:19girl he likes the
21:22most okay every day the boy every boy
21:27any boy will go and stand in front of
21:30the balcony of the girl he likes the
21:32most
21:34okay the girl the each day a
21:38girl can
21:40either uh tell a boy standing in front
21:43of the balcony there could be more than
21:44one boy standing in front of a girl's
21:46balcony the girl can tell the boy come
21:50back next day or
21:53reject okay once the girl says reject to
21:57a boy the boy crosses that girl off from
22:01his list of
22:02possibilities and never goes back to
22:05that balcony ever okay this
22:09continues until every girl has exactly
22:13one boy standing in front of the
22:17balcony
22:19okay when there is exactly one boy
22:22standing in front of each girl's balcony
22:25then the algorithm terminates
22:28this is the stable matching
22:31algorithm and let's see if this
22:34works and initially all girls are in the
22:37boy's
22:38list he'll keep the boy will keep
22:41crossing a girl out once the girl says
22:44reject okay fine let's look at the
22:47iterations now let's apply this
22:49algorithm on the example which we have
22:53got so this is our preference lists
22:56remember so let's see this is day one
22:59what's going to
23:02happen uh every boy will go and stand in
23:05front
23:06of uh the balcony of the girl he likes
23:09the most
23:11okay
23:13so one whom does one like the most boy
23:17one boy one likes C so boy one goes and
23:21stands in front of C's
23:23balcony whom does two like the most a so
23:27two goes and stands in front of A's
23:29balcony whom does three like the most
23:32D where does d go d goes and stand in
23:36front
23:37of uh sorry three goes and stand in
23:39stands in front of D's
23:41balcony what about boy four boy four
23:45likes a the most and boy five also likes
23:48a the most so both of them again go and
23:50stand in front of A's
23:53balcony okay so this is how it's
23:56operating great
23:59now two knows that both two four and
24:03five prefer her the
24:06most right girl girl a knows that boys 2
24:114 and five prefer her the
24:15most okay now look at girl A's
24:18preference preference ordering whom does
24:21girl a prefer the most amongst these
24:23boys 2 4 and five well clearly girl a
24:28prefers five the
24:30most
24:32okay so girl a knows that five is
24:35available to her then why should she
24:38bother about boys 2 and four so what
24:42will she say she will say reject to 2
24:44and
24:45four okay she will say reject to 2 and
24:50four so two and four will now what will
24:54what will two and four do boys two and
24:56four what will they do
24:58well two and four will cross a out of
25:01their list so like this a is out of
25:04their list now Mark Mark a
25:07red
25:08h fine now day two comes in day two
25:12again every boy goes and stands in front
25:15of the balcony of the girl he likes the
25:18most in his
25:20list in his
25:22list so again one goes in front of C's
25:26balcony right two will now go in front
25:29of B's balcony three will now go in
25:32front of C's balcony four will now go in
25:35front of C's balcony right five in front
25:38of a so this is what happens
25:43now
25:46right now C is having two two boys
25:51standing in front of her balcony one and
25:54four now whom does she like more well C
25:58like four the most so she she and she
26:02knows that
26:05uh right now she is the most preferred
26:09girl in boy Four's list so why should C
26:13bother about boy one why should girl C
26:17bother about boy one so girl C rejects
26:21boy
26:22one
26:24okay so boy one now crosses out see from
26:29his list so these are my cross outs
26:33now
26:35right
26:37fine now again now day three comes
26:40everybody goes every boy goes and stands
26:43in front of the balcony of the girl he
26:45likes the most in his
26:48list okay remember if you look at boy
26:52one's list C is not there anymore so
26:55where will he go he will go and stand in
26:57front of B B's balcony two again will go
27:00and stand in front of B's balcony
27:04right so 1 and two both go and stand in
27:07front of B's
27:08balcony right you can you can see that
27:11three will go in front of D's balcony
27:13four will go and stand in front of C's
27:15balcony because a is not there in Four's
27:18list
27:19anymore five again will go and stand in
27:21front of A's balcony so this is how the
27:24balconies look like
27:26now now B has two men standing in front
27:29of her balcony one and
27:32two look at B's ordering B's preference
27:35ordering between 1 and two who does B
27:39prefer well B clearly prefers two more
27:42than
27:43one right and B knows that right now she
27:48is she tops in the preference list of
27:53two if that's the case why should she
27:56bother about boy one so she rejects boy
28:01one so B will reject boy one and boy one
28:05in turn will cross out B from his
28:09list this is the updated list
28:12now great what happens next day where
28:14will boy one go in front of the most
28:17preferred girl of his list these have
28:19been crossed out so boy one goes in goes
28:23and stands in front of E's balcony two
28:25goes and stands in front of B's balcony
28:28three in front of D's balcony four in
28:30front of C's balcony five in front of
28:32A's
28:33balcony this is what we
28:36have okay now we have the terminating
28:39condition right we have one boy standing
28:43in front
28:44of uh or every balcony has exactly one
28:48boy standing in
28:50front and remember this was my condition
28:53for termination of the stable matching
28:55algorithm so this is where the algorithm
28:57terminates
29:00fine now is this a stable match is this
29:03match which we have got now is this
29:05stable remember we had got a similar
29:07match using the greedy algorithm which
29:08turned out to be unstable because we
29:10could find a rogue
29:12couple but this match which we have got
29:15let's see if this is
29:17stable and the answer
29:19is if you take a look at the preference
29:23of the boys and the preference of the
29:24girls and if you take a look at this
29:26matching it indeed turns out to be
29:29stable you will not be able to find any
29:32Rogue couple in this
29:35matching I will urge all of you to take
29:38a little pause and work it out to
29:40yourself try to find a rogue couple and
29:43you will see that you
29:44can't
29:48okay now the question is fine we have
29:50proposed an
29:52algorithm now uh let's look at some
29:54desirable properties of the stable
29:56matching algorithm and this in turn will
30:00U kind of make it clear that this
30:04algorithm Works generally under any
30:07situation for any preference orderings
30:10if the preference orderings can be uh if
30:13if it's a bipartite
30:16structure okay so the first first result
30:20the stable matching algorithm
30:21necessarily terminates it will terminate
30:23at some
30:24point what is the proof for that
30:28it's very
30:30simple in fact it terminates in less
30:33than equal to n² + 1 days that's the
30:35upper
30:36bound why what is the logic what is
30:40happening in this algorithm every day at
30:43least one boy is being crossed
30:47out or or in other words one boy crosses
30:50out a girl from his list when every day
30:54one girl is crossed
30:56out from some boys
30:59list so every day there is a cross out
31:03now there are n boys and N girls and so
31:07every boy has a list which which has a
31:09cardinality n of size n so how many
31:13total cross outs are possible at
31:16Max well at Max n Square cross outs are
31:20possible Right which means on the n² + 1
31:24at day the algorithm will necessarily
31:26terminate
31:29okay so this is the proof that the
31:31stable matching algorithm will
31:33necessarily terminate if we uh have a
31:36bunch of boys and girls and the
31:38preference orderings and if we apply the
31:40algorithm it will necessarily terminate
31:42at some point after a finite amount of
31:45time and the upper upper bound is n sare
31:48+
31:491 upper bound of
31:53time what is the next one everybody gets
31:56married or paired so this is the
31:58definition of a perfect matching
31:59remember in perfect matching everybody
32:01gets
32:03paired uh in our example or we have made
32:07a simplifying assumption at the start of
32:08the lecture that uh the set of the the
32:11the cardinality of the two sets which
32:13are being matched are
32:15equal so let's prove this let's say boy
32:19B has not been uh is not married at the
32:24end okay any boy I I'm naming him me
32:29which means B has been rejected by every
32:32girl
32:34right which means every girl is already
32:38married but if every girl is married
32:40every boy is
32:41married because the cardinality of the
32:44two sets are equal so if every girl is
32:47married every girl is married to one boy
32:49at least which means every boy is
32:51married which means B is also married so
32:54which which leads to a contradiction
32:56that b is not married okay so if the
32:58cardinality of the two sets are equal
33:00this is not
33:03possible and finally the last
33:06one which is that uh it this algorithm
33:10necessarily produces a stable match we
33:13have seen it does right we have seen it
33:16does in the example which we worked
33:20out but what is the intuitive uh
33:22explanation that it always does it's the
33:26following right it's it's a pretty
33:28intuitively easy proof to think about
33:31let us say it does not let us say we
33:33have applied stable matching and we
33:35actually end up getting a rogue couple
33:37let's say the Rog couple is called
33:38Johnny and
33:39Amber okay now if it is a rogue couple
33:42it means it is not a couple in the
33:44matching right now
33:46right okay now how come Johnny and Amber
33:50are not couple right now in the St in
33:53this in the match which we have got
33:55either case one is either Amber rejected
33:59Johnny right if Amber rejected Johnny it
34:03means Amber must have
34:06had a boy more preferred than Johnny
34:11standing in front of her balcony only
34:13then Amber would have rejected
34:17journy and the final Choice which Amber
34:20got was definitely
34:23somebody who was preferred to
34:26Journey which which means Johnny and
34:29Amber cannot be a rogue couple so it's a
34:33contradiction what is case two case 2 is
34:35Johnny never went and serated Amber that
34:38is Johnny never went and stood in front
34:39of Amber's balcony what does this
34:43mean now when is a boy going to a
34:46balcony of a
34:49girl when is a boy not going in front of
34:52balcony of a
34:53girl when he's not being rejected by a
34:56girl
34:58whom she whom he prefers more than the
35:01girl uh under whose balcony he has not
35:04been to right so which means if Johnny
35:07has not been under the balcony of Amber
35:10it simply means that Johnny has not been
35:12rejected by some girl whom he prefers
35:16more than
35:18Amber if he has not been rejected by a
35:20girl whom he prefers more than
35:23amember which means Johnny's current
35:25partner whom he has been matched with by
35:28the
35:30algorithm he prefers that person more
35:32than
35:33am which means Johnny does not prefer
35:36Amber over his current
35:38partner which means Johnny and Amber
35:40again does not form a rogue couple right
35:45so we see again we prove it by
35:47contradiction that uh formation of a
35:50rogue couple or occurrence of a rogue
35:52couple is impossible once we apply
35:55stable matching algorithm
35:59okay
36:00great so we have learned uh
36:04uh the stable matching algorithm in in
36:07this lecture so we have learned two
36:09different kinds of things we have had
36:10two different uh uh exposures in the
36:13first two lectures in the first lecture
36:16we talked about network data and we
36:18tried to interpret two different
36:20situations one in history one in
36:23finance in this lecture we have looked
36:26at something else we have looked at
36:27stable matching algorithm in the next
36:30two or three lectures I will deal with
36:33something completely
36:35different I will talk about modeling of
36:39uncertainty in artificial intelligence
36:41training an agent to behave optimally in
36:45uncertain situations is a key
36:48thing so is true in
36:51finance so in the next two lectures I
36:54will delve a little more in in finance
36:58and talk about the idea of hedging and
37:01risk
37:02management see you in the next lecture

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