Dating%20Scenario

1
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
  all the girls
• The lists have no ties
• Question How to pair them off?

2
Stable Marriage

• Would like a pairing that is stable
• No pairs (g1,b1) and (g2,b2) exist such that g1
  prefers b2 over b1 and b2 prefers g1 over g2
• The pair (g1,b2) will be referred to as a rogue
  couple
• A stable marriage is a pairing that allows no
  possibility of rogue couples

3
Example
Albert Laura Nancy Karen Mimi
Brad Mimi Nancy Karen Laura
Chuck Mimi Karen Nancy Laura
David Karen Laura Mimi Nancy
Karen Albert Chuck David Brad
Laura Brad Chuck David Albert
Mimi Brad David Albert Chuck
Nancy Chuck Albert David Brad
4
Stable Marriage Problem

• Given preference lists, determine a stable
  pairing
• A more fundamental question
• Does a stable pairing always exist??

5
Instructive Variant Bisexual case

• Consider 4 participants A, B, C, and D
• A B C D
• B C A D
• C A B D
• D does not matter
• Any pairing is unstable
• say (A,B) and (C,D)
• C prefers B over D and B prefers C over A

6
Existence of Stable Pairing

• For the heterosexual case, stable pairing always
  exists!
• Will show that a simple algorithm the
  traditional marriage algorithm (TM) always
  yields a stable marriage

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TM Algorithm (Overview)

• The algorithm proceeds in rounds.
• In each round
• Each boy proposes to one girl
• Each girl rejects all but one boy
• At the end, a stable pairing will result

8
TM Algorithm

• In each round
• Each boy proposes to his most-preferred girl, who
  has not rejected him
• Each girl considers the proposals in the round
  rejects all but the most-preferred proposal
• Repeat until no rejections occur
• Each girl marries the sole proposer in last round
• We have a stable marriage!

9
Properties of TM

• Improvement Once a girl has a suitor, she will
  always have a suitor furthermore, the ranking of
  the suitor will never decrease
• Matching No boy can be rejected by all girls
• Termination TM always terminates in at most
  rounds
• Stability The final pairing is stable

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Improvement

• Suppose a girl G has at least one proposer in
  round i.
• Suppose B is the proposer not rejected by G in
  round i
• G has a suitor B
• B will repeatedly propose to G until G rejects B
  and takes a better suitor (or marries B)
• Thus, once G has a suitor, she always has a
  suitor
• Ranking of the suitor never decreases

11
Matching

• Suppose boy B is rejected by all girls
• Since B has proposed to all girls, each of them
  has had a suitor at some point
• This implies they all have suitors now (by the
  improvement lemma)
• But this is a contradiction, since we have n
  girls and n-1 suitors (since B is not one)

12
Termination

• A boy cannot rejected by a girl twice
• There are a total of possible rejections
• Each round until the end there is at least
  one rejection
• So at most rounds before TM terminates

13
Stability

• Suppose the resultant pairing is not stable
• (G1, B1) and (G2, B2) are couples
• B1 prefers G2 over G1 and G2 prefers B1 over B2
• Can this happen?
• B1 must have proposed to G2 before G1
• G2 rejected B1 for some other boy B3 who was
  higher on G2s list
• The final suitor for G2, B2, is at least as high
  on G2s list as B3
• So G2 prefers B2 over B1

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Who Fares Better in TM?

• Boys Start from the top of their list
• Girls The ranking of their suitor progressively
  increases
• In final solution
• Each boy is paired with the highest ranked girl
  on his list that he can conceivably get in a
  stable world
• Each girl is paired with the lowest ranked boy on
  her list that she can conceivably get in a stable
  world

15
Stable Marriage in Practice

• Matching Residents to Hospitals
• Since WWII, medical school graduates have been
  matched to hospitals (as residents) using a
  stable marriage algorithm
• College Admissions