Matching Problems. Roberto Lucchetti. Politecnico di Milano

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Politecnico di Milano

Background setting Problems introduced in 1962 by Gale and Shapley for the study of two sided markets: 1) workers & employers; 2) interns & hospitals; 3) students & universities; 4) women & men; 5) Basic question: what is the best definition of stable placements?

First basic version WOMEN & MEN Two groups A and B. Each element of A (B) has a ranking on the elements of B (A). Searching for pairs One to one matching Further initial assumptions 1) The two groups have the same number of elements; 2) All elements rank all elements of the other group.

Basic definition Definition Let X be a set. A (strict) preference relation on X is a binary relation fulfilling, for all x, y, z X : 1) if x y either x y or y x (completeness); 2) x x (irreflexivity); 3) x y y z imply x z (transitivity). Meaning 1) Each pair is compared; 2) Preferences are strict; 3) Preferences are coherent.

More definitions Definition A matching problem is given by: 1) a natural number n (the common cardinality of two distinct sets A and B, whose elements are called women and men); 2) a set of preferences such that each woman has a preference relation over the set of men and conversely. Definition A matching is a bijection between the two sets.

An example A = {Anna, Giulia, Maria} B = {Bob, Frank, Emanuele} A matching [(Anna, Emanuele), (Giulia, Bob), (Maria, Frank)].

Stable matching Definition A man m and a woman W object to the matching Λ if m and W both prefer each other to the person paired to them in the matching Λ. For instance: Λ = {(m, W ), (b, Z),... } and b W m W b Z. Definition A matching Λ is called stable provided there is no pair woman man objecting to Λ.

Example M = {a, b, c, d}, W = {A, B, C, D}. Preferences men: D a C a B a A, B b A b C b D D c A c C c B, A d D d C d B. Preferences women: b A a A c A d, b B d B a B c b C d C c C a, b D c D d D a. Two matchings: Λ = {(a, A), (b, C), (c, B), (d, D)} Ω = {(a, C), (b, B), (c, D), (d, A)} Λ unstable (a,b) Ω stable.

Existence of stable matching Theorem Every matching problem admits a stable matching

Proof Constructive proof: algorithm night by night 1) Stage 1a) Every woman visits, the first night, her most preferred choice. Stage 1b) Every man chooses among the women he finds in front of his house (if any). If every woman is matched, the algorithm ends, otherwise go to stage 2 2) Stage 2a) Every woman dismissed at the previous stage visits her second choice. Stage 2b) Every man chooses among the women he finds in front of his house (if any), he dismisses the woman already in the house, if someone at the door is better for him. If every woman is matched, the algorithm ends, otherwise go to stage k 3) Stage ka) Every woman dismissed at the previous stage visits her choice after the man dismissing her. Stage kb) Every man chooses among the women he finds in front of his house (if any), he dismisses the woman already in the house, if someone at the door is better for him. If every woman is matched, the algorithm ends, otherwise go to stage k. Claim. The algorithm ends and the resulting matching is stable

Preliminary remarks Easy to see 1) The women go down with their preferences along the algorithm; 2) The men go up with their preferences along the algorithm; 3) If a man is visited at stage r, then from stage r + 1 on he will never be alone; 4) The algorithm generically provides two matchings.

Proof: continued Fact 1 The algorithm ends, and every man is matched to a woman. Actually every woman can at most visit n men, and this happens to every woman. Thus at most n 2 days are needed. Noticing that the first night all women are involved, the estimate n 2 n + 1 1 is clear. Moreover, every man is visited at some stage: all women like better to be paired than to remain alone (Under the assumption of equality of the two groups and completeness of preferences). By second remark above once visited no man remains alone. Fact 2 No woman can be part of an objecting pair for the resulting matching. Consider the woman W : she cannot be part of an objecting pair with a man she did not visit: she is married to a man preferred to all of them. She cannot be part of an objecting pair with a man m she already visited either: dismissed in favor of another woman, by transitivity she cannot be preferred to the woman matched to m. The matching built by the algorithm is stable. 1 This estimate is not sharp.

Example revisited Preferences men: D a C a B a A, B b A b C b D D c A c C c B, A d D d C d B. Preferences women: b A a A c A d, b B d B a B c b C d C c C a, b D c D d D a. Men visiting: Women visiting {(a, C), (b, B), (c, D), (d, A)}. {(a, A), (b, B), (c, D), (d, C)}.

One more example Preferences men: A a B a C, C b A b B, B c A c C. Preferences women: c A a A b, b B a B c, a C b C c Men visiting: {(a, A), (b, C), (c, B)}. Women visiting: {(c, A), (b, B), (a, C)}. One more? {(a, B), (b, C), (c, A)}.

Some numbers (matching) =n! (stable matching) =? Usually not many! However there is a method allowing to built many stable matching. For instance: n = 8 269, n = 16 195472 n = 32 104310534400.

Comparing matching Let us consider two matching and Θ. Definition Write m Θ if for every man it happens that either he is associated to the same woman in the two matchings or he is associated to a preferred woman in. Write w Θ if for every woman it happens that either she is associated to the same man in the two matchings or she is associated to a preferred man in. Remarks m ( w ) reflexivity; m ( w )Θ Θ m ( w )Λ imply m ( w )Λ transitivity; They are not complete ordering.

Women versus men Theorem Let and Θ be stable matching. Then m Θ if and only if Θ w. Suppose m Θ, and (a, A), (b, A) Θ. Must show b A a. Suppose in (a, F ) Θ. Since m Θ Moreover and so A a F. {(a, F ), (b, A)} Θ b A a.

Ordering stable matching Theorem Let Λ m (Λ w ) be the men (women) visiting matching and let Θ be another stable matching. Then Λ m m Θ m Λ w Λ w w Θ w Λ m. Women visiting is the best for the women.

The proof Proving that a woman cannot be rejected by a man available to her. Induction on the days of visit First day. Suppose A is rejected by a in favor of B and that there is a stable set such that {(a, A), (b, B)}. Then implying B a A b B a. Impossible! B is visiting a the first day!

The proof continued Suppose A was not rejected by an available man the days 1,..., k 1. See that A cannot be refused by an available man the day k. Suppose A is rejected by a in favor of B in day k and that there is a stable set such that {(a, A), (b, B}. Then implying B a A b B a. This means B had visited a some day before and was rejected, impossible!

Proof finished See that Λ w w Θ for every stable Θ. Let (a, A) Λ w. Must show that if (b, A) Θ then a A b. Suppose the contrary. Then A visited b before a and was rejected. Impossible! To conclude, use previous theorem and symmetry between men and women.

Further result Definition Let, Θ be two matching. Define a new matching w Θ as the matching where each woman is paired to the preferred man between the two paired to her in and Θ Theorem Let, Θ be two stable matching. Then w Θ is stable. Mathematically, w provides a lattice structure to the set of stable matchings.

An extension One can consider the case when the number of men and women are different, for instance men are more. Definition A matching is a function assigning to every man an element of the set W {single} with every woman associated to a man. Assuming that for men remaining alone is worse than being paired to any woman, then the woman A and the man a object to a matching if a is either single or paired to a woman for him worse than A, and A loves more a than the man paired to her in the matching. A matching is stable if there are no objecting pairs. Theorem The set of the men which are not married remains the same on all stable matchings.

A key issue Is it possible that people lie on their true preferences? It is intuitive, and true, that in the case of the women visiting, they do not have incentive to lie. But what about men? Examples show that in some cases for a man could be convenient to lie.

Further extensions 1) Getting married, but not at any cost; 2) Polygamous matching; 3) Unisexual matching. 1) adaptation of the idea of stable matching, the algorithm works; 2) adaptation of the idea of stable matching, the algorithm works; 3) a stable matching need not to exist.