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Lecture 23: Stable Marriage ( Based on Lectures of Steven Rudich of CMU and Amit Sahai of Princeton). Shang-Hua Teng. Dating Scenario. There are n “men” and n “women” Each woman has her own ranked preference list of ALL the men Each man has his own ranked preference list of ALL the women
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Lecture 23:Stable Marriage(Based on Lectures of Steven Rudich of CMU and Amit Sahai of Princeton) Shang-Hua Teng
Dating Scenario • There are n “men” and n “women” • Each woman has her own ranked preference list of ALL the men • Each man has his own ranked preference list of ALL the women • The lists have no ties
3,2,5,1,4 1,3,4,2,5 4,3,2,1,5 1,2,5,3,4 3,5,2,1,4 1,2,4,5,3 1 4 1 3 2 5 5,2,1,4,3 2 4,3,5,1,2 3 1,2,3,4,5 4 2,3,4,1,5 5
The Matching (Marriage) Problem Question: How do we pair them off? Which criteria come to mind?
There is more than one notion of what constitutes a “good” pairing. • Maximizing the number of people who get their first choice • Maximizing average satisfaction • Maximizing the minimum satisfaction
Couple of “Mutually Cheating Hearts” • Suppose we pair off all the men and women. Now suppose that some man and some woman prefer each other to the people they married. They will be called a couple of MCHs.
Stable Pairings • A pairing of men and women is called stable if it contains no couple of MCHs.
Stability is Primary. • Any list of criteria for a good pairing must include stability. (A pairing is doomed if it contains a shaky couple.) • Any reasonable list of criteria must contain the stability criterion.
The study of stability will be the subject of the entire lecture. • We will: • Analyze an algorithm that looks a lot like dating in the early 1950’s • Discover the naked mathematical truth about which sex has the romantic edge • Learn how the world’s largest, most successful dating service operates
How do we find a stable pairing? Wait! There is a more primary question!
How do we find a stable pairing? How do we know that a stable pairing always exists?
A “Traditional” Marriage Algorithm Choose the best among those who propose, “get a commitment” Propose to the best who is still available
Traditional Marriage Algorithm • For each day each single man does the following: • Morning • Each bachelorette sits in her favorite coffee house • Each single men proposes to the best bachelorette to whom he has not yet proposed • Afternoon (for those bachelorettes with at least one suitor) • To today’s best suitor: “Maybe, come back tomorrow, and ask me again” • To any others: “Not me, but good luck” • Evening • Each single man is ready to propose next bachelorette on his list The day WILL COME that no bachelor is rejected Each bachelorette marries the last bachelor to whom she said “maybe”
Does the Traditional Marriage Algorithm always produce a stable pairing?
Does the Traditional Marriage Algorithm always produce a stable pairing? Wait! There is a more primary question!
Does the TMA work at all? • Does it eventually stop? • Is everyone married at the end?
Theorem: The TMA always terminates in at most n2 days • Each day that at least one Bachelor gets a “No”. • There can be at most n2 “No”s.
Does the TMA work at all? • Does it eventually stop? • Is everyone married at the end?
Improvement Lemma: If a bachelorette has a committed suitor, then she will always have someone at least as good the next day (and on the final day). • She would only let go of him in order to “maybe” someone better • She would only let go of that guy for someone even better • She would only let go of that guy for someone even better • AND SO ON . . . . . . . . . . . . .
Corollary: Each bachelorette will marry her absolute favorite of the bachelors who proposed to her during the TMA
Lemma: No bachelor can be rejected by all the bachelorettes Proof by contradiction. Suppose Bob is rejected by all the women. At that point: Each women must have a suitor other than Bob (By Improvement Lemma, once a woman has a suitor she will always have at least one) The n women have n suitors, Bob not among them. Thus, there must be at least n+1 men! Contradiction
Great! We know that TMA will terminate and produce a pairing.But is it stable?
Luke Bob Alice Mia Theorem: The pairing produced by TMA is stable. • Proof by contradiction:Suppose Bob and Mia are a couple of MCHs. • This means Bob likes Mia more than his partner, Alice. • Thus, Bob proposed to Mia before he proposed to Alice. • Mia must have rejected Bob for someone she preferred. • By the Improvement lemma, she must like her parnter Luke more than Bob. Contradiction!
Opinion Poll Who is better off in traditional dating, the men or the women?
Forget TMA for a moment • How should we define what we mean when we say “the optimal woman for Bob”? Flawed Attempt: “The woman at the top of Bob’s list”
The Optimal Match A man’s optimal match is the highest ranked woman for whom there is some stable pairing in which they are matched She is the best woman he can conceivably be matched in a stable world. Presumably, she might be better than the woman he gets matched to in the stable pairing output by TMA.
The Pessimal Match A man’s pessimal match is the lowest ranked woman for whom there is some stable pairing in which the man is matched to. She is the least ranked woman he can conceivably get to be matched to in a stable world.
Dating Dilemmas • A pairing is man-optimal if everyman gets his optimalmatch. This is the best of all possible stable worlds for every man simultaneously. • A pairing is man-pessimal if everyman gets his pessimalmatch. This is the worst of all possible stable worlds for every man simultaneously.
Dating Dilemmas • A pairing is woman-optimal if every woman gets her optimalmatch. This is the best of all possible stable worlds for every woman simultaneously. • A pairing is woman-pessimal if every woman gets her pessimal match. This is the worst of all possible stable worlds for every woman simultaneously.
The Naked Mathematical Truth! The Traditional Marriage Algorithm always produces a man-optimal, and woman-pessimal pairing.
Theorem: TMA produces a man-optimal pairing • Suppose not: i.e. that some man gets rejected by his optimal match during TMA. • In particular, let’s sayBobis thefirst manto be rejected by his optimal match Mia: Let’s say she said “maybe” toLuke, whom she prefers. • Since Bobwas the only man to be rejected by his optimal match so far, Lukemust likeMiaat least as much as his optimal match.
Mia Luke We are assuming that Mia is Bob’s optimal match Mia likes Luke more than Bob. Luke likes Mia at least as much as his optimal match. • We now show that any pairing S in which Bob marries Mia cannot be stable (for a contradiction). • Suppose S is stable: • LukelikesMiamore than his partner in S • LukelikesMia at least as much as his best match, but he is not matched to Mia in S • MialikesLukemore than her partnerBobin S Contradiction!
We are assuming that Mia is Bob’s optimal match.Mia likes Luke more than Bob. Luke likes Mia at least as much as his optimal match. • We’ve shown that any pairing in which Bob marries Mia cannot be stable. • Thus, Miacannot be Bob’s optimal match(since he can never marry her in a stable world). • So Bob never gets rejected by his optimal matchin the TMA, and thus the TMA is man-optimal.
Alice Luke Theorem:The TMA pairing is woman-pessimal. • We know it is man-optimal. Suppose there is a stable pairing S where some woman Alice does worse than in the TMA. • Let Luke be her partner in the TMA pairing. Let Bob be her partner in S. • By assumption, AlicelikesLukebetter than her partnerBob in S • LukelikesAlicebetter than his partner in S • We already know that Alice is his optimal match ! Contradiction!
The largest dating service in the world uses a computer to run TMA !
“The Match”:Doctors and Medical Residencies • Each medical school graduate submits a ranked list of hospitals where he/she wants to do a residency • Each hospital submits a ranked list of newly minted doctors that it wants • A computer runs TMA (extended to handle polygamy) • Until recently, it was hospital-optimal
2,3,4 3,1,4 1 2 1,2,4 *,*,* 3 4 An Instructive Variant:Team Projects
3,1,4 2 *,*,* 4 An Instructive Variant:Team Project 2,3,4 1 1,2,4 3
3,1,4 2 *,*,* 4 An Instructive Variant:Team Project 2,3,4 1 1,2,4 3
3,1,4 2 *,*,* 4 An Instructive Variant:Team Project 2,3,4 1 1,2,4 3
3,1,4 2 *,*,* 4 No Stable Teaming Possible 2,3,4 1 1,2,4 3
Stability • Amazingly, stable pairings do always exist in the bipartite relations. • Insight: We must make sure our arguments do not apply to the class project case, since we know all arguments must fail in that case.
REFERENCES • D. Gale and L. S. Shapley, College admissions and the stability of marriage, American Mathematical Monthly 69 (1962), 9-15 • Dan Gusfield and Robert W. Irving, The Stable Marriage Problem: Structures and Algorithms, MIT Press, 1989
History: When Markets fail • 1900 • Idea of hospitals having residents (then called “interns”) • Over the next few decades • Intense competition among hospitals for an inadequate supply of residents • Each hospital makes offers independently • Process degenerates into a race. Hospitals steadily advancing date at which they finalize binding contracts
History: When Markets fail • 1944 Absurd Situation. Appointments being made 2 years ahead of time! • All parties were unhappy • Medical schools stop releasing any information about students before some reasonable date • Did this fix the situation?
History: When Markets fail • 1944 Absurd Situation. Appointments being made 2 years ahead of time! • All parties were unhappy • Medical schools stop releasing any information about students before some reasonable date • Offers were made at a more reasonable date, but new problems developed