15 251 Some Great Theoretical Ideas in Computer Science for CMU Student Crush Script http 1982087 com crush html The Mathematics Of 1950 s Dating Who wins The Battle of The Lecture 10Sexes February 12 2009 WARNING This lecture contains mathematical content that may be shocking to some students Dating Scenario There are n boys and n girls Each girl has her own ranked preference list of all the boys Each boy has his own ranked preference list of the girls The lists have no ties Question How do we pair them off 3 2 5 1 4 3 5 2 1 4 1 1 5 2 1 4 3 1 2 5 3 4 2 2 4 3 5 1 2 4 3 2 1 5 3 3 1 2 3 4 5 1 3 4 2 5 4 4 2 3 4 1 5 1 2 4 5 3 5 5 More Than One Notion of What Constitutes A Good Pairing Maximizing total satisfaction Hong Kong and to an extent the USA Maximizing the minimum satisfaction Western Europe Minimizing maximum difference in mate ranks Sweden Maximizing people who get their first choice Barbie and Ken Land We will ignore the issue of what is best Rogue Couples Suppose we pair off all the boys and girls Now suppose that some boy and some girl prefer each other to the people to whom they are paired They will be called a rogue couple Why be with them when we can be with each other What use is fairness if it is not stable Any list of criteria for a good pairing must include stability A pairing is doomed if it contains a rogue couple Stable Pairings A pairing of boys and girls is called stable if it contains no rogue couples Stable Pairings A pairing of boys and girls is called stable if it contains no rogue couples 3 2 1 3 2 1 1 1 2 1 3 1 2 3 2 2 3 1 2 3 2 1 3 3 The study of stability will be the subject of the entire lecture We will Analyze various mathematical properties of an algorithm that looks a lot like 1950 s dating Discover the naked mathematical truth about which sex has the romantic edge Learn how the world s largest most successful dating service operates Given a set of preference lists how do we find a stable pairing Wait We don t even know that such a pairing always exists Better Question Does every set of preference lists have a stable pairing Idea Allow the pairs to keep breaking up and reforming until they become stable Can you argue that the couples will not continue breaking up and reforming forever 2 3 4 An Instructive Variant Bisexual Dating 1 3 1 4 2 1 2 4 3 4 Insight Any proof that heterosexual couples do not break up and re form forever must contain a step that fails in the bisexual case If you have a proof idea that works equally well in the hetero and bisexual versions then your idea is not adequate to show the couples eventually stop The Traditional Marriage Algorithm Worshipping Males Female String The Traditional Marriage Algorithm For each day that some boy gets a No do Morning Each girl stands on her balcony Each boy proposes to the best girl whom he has not yet crossed off Afternoon for girls with at least one suitor To today s best Maybe return tomorrow To any others No I will never marry you Evening Any rejected boy crosses the girl off his list If no boys get a No each girl marries boy to whom she just said maybe 3 5 2 1 4 3 2 5 1 4 1 1 5 2 1 4 3 1 2 5 3 4 2 2 4 3 5 1 2 4 3 2 1 5 3 3 1 2 3 4 5 1 3 4 2 5 4 4 2 3 4 1 5 1 2 4 5 3 5 5 Does Traditional Marriage Algorithm always produce a stable pairing Wait There is a more primary question Does TMA Always Terminate It might encounter a situation where algorithm does not specify what to do next e g core dump error It might keep on going for an infinite number of days Improvement Lemma If a girl has a boy on a string then she will always have someone at least as good on a string or for a husband 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 girl will marry her absolute favorite of the boys who visit her during the TMA Lemma No boy can be rejected by all the girls Proof by contradiction Suppose boy b is rejected by all the girls At that point Each girl must have a suitor other than b By Improvement Lemma once a girl has a suitor she will always have at least one The n girls have n suitors and b is not among them Thus there are at least n 1 boys Contradictio n Theorem The TMA always terminates in at most n2 days A master list of all n of the boys lists starts with a total of n x n n2 girls on it Each day at least one boy gets a No so at least one girl gets crossed off the master list Therefore the number of days is bounded by the original size of the master list Great We know that TMA will terminate and produce a pairing But is it stable Theorem The pairing T produced by TMA is stable g I rejected you when you came to my balcony Now I ve got someone better b g Opinion Poll ff o r e t t l e a b n s i o s i t y o i o h d b a W r t e in g th rls n e gi i t da r th o Forget TMA For a Moment How should we define what we mean when we say the optimal girl for boy b Flawed Attempt The girl at the top of b s list The Optimal Girl A boy s optimal girl is the highest ranked girl for whom there is some stable pairing in which the boy gets her She is the best girl he can conceivably get in a stable world Presumably she might be better than the girl he gets in the stable pairing output by TMA The Pessimal Girl A boy s pessimal girl is the lowest ranked girl for whom there is some stable pairing in which the boy gets her She is the worst girl he can conceivably get in a stable world Dating Heaven and Hell A pairing is male optimal if every boy gets his optimal mate This is the best of all possible stable worlds for every boy simultaneously A …
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