CS 411 Fall 2025  >  Outline for November 14, 2025

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CS 411 Fall 2025
Outline
for November 14, 2025

Outline

The Stable Marriage Problem [L 10.4]

• The Problem
  • Given two sets of equal size, find a matching between them that is optimal according to certain criteria. Usually stated in terms of “men” and “women”, but there are many applications.
  • Idea
    • Given:
      • Two disjoint sets of equal size (\(n\)): a set of men and a set of women.
      • For each element of each set, an ordering (best to worst) of the elements of the other set. We think of this as how the man/woman rates or prefers members of the other gender.
    • A marriage matching is a collection of pairs, each containing one woman and one man, such that each woman or man lies in exactly one pair.
    • A marriage matching is unstable if there are a man & a woman, not paired together, each of whom prefers the other to their current partner. Otherwise, the marriage matching is stable.
  • Two Questions
    1. Does a stable marriage matching always exist? (Answer turns out to be yes.)
    2. How to find one? This is the Stable Marriage Problem.
• Algorithm: Description
  • Input is as above.
  • We keep track of a changing set of pairs, each with one man and one woman. A man or a woman that is not currently in a pair is free. A man & woman that form a pair are said to be matched. No man or woman will ever be matched with more than one person at the same time.
  • Procedure
    • All men, women are initially free.
    • While there are free men/women:
      • Choose a free man \(m\).
      • Find the woman \(w\) that is highest in \(m\)’s ratings, among all those who have not yet said “no” to \(m\). (Does such a woman necessarily exist? See the Note, below.*)
      • \(m\) proposes to \(w\)—regardless of whether \(w\) is free or matched.
      • If \(w\) is free, then \(w\) says “yes” to \(m\); \(w\) and \(m\) are now matched.
      • If \(w\) is matched and prefers her current partner to \(m\), then \(w\) says “no” to \(m\); the pairs are unchanged.
      • If \(w\) is matched and prefers \(m\) to her current partner, then \(w\) says “no” to her current partner and “yes” to \(m\); the existing pair including \(w\) is dissolved, and \(w\) and \(m\) are now matched.
    • Return the set of pairs. END.
    • *Note. If \(m\) is a free man, will there always be a woman who has not yet said “no” to \(m\)? Since there is a free man, there must be a free woman. Because of the way the algorithm works, a woman who is free has never been proposed to, so she has never never said “no” to \(m\). So, yes, there will always be a woman who has not yet said “no” to \(m\).
• Algorithm: Termination
  • Each time we process a free man \(m\), one of two things happens.
    1. The number of pairs increases (if \(w\) is free, she will say “yes” to \(m\), forming a new pair and not dissolving any existing pairs).
    2. The number of “no” responses increases (if \(w\) is matched, she will say “no” either to \(m\) or to her current partner).
    Neither of these two ever reverses, and both have limits. Thus, the algorithm must eventually terminate.
• Algorithm: Correctness
  • Proof by contradiction: if the algorithm terminates without a stable marriage matching found, then we can argue that the algorithm did not run correctly.
  • Since algorithm always ends with stable marriage matching found, one must always exist.
• Algorithm: Analysis
  • Time Efficiency
    • Store info needed by each man in an array, and similarly for the women. Refer to a man/woman by their array index.
    • Each man has a queue of women, best to worst.
    • Each woman has a look-up table: \(\text{man}\to\text{rating}\); this can be an array. She also needs to know who she is matched with, if anyone: an array index with possible no one value.
    • Store the set of free men in a sequence with constant-time access and deletion at one end, for example, a resizable array, using the end, or a singly linked list, using the beginning.
    • With data in this form, checking for termination and choosing a free man is constant-time.
    • Processing a free man is also constant-time. Note that, if a man’s proposal is accepted by a woman who is currently matched, then we can just replace the current free man with the newly rejected man in the free-men data structure.
    • Each man proposes to each woman at most once. So the inside of the loop runs at most \(\Theta(n^2)\) times.
    • Preprocessing—setting up the data—requires \(\Theta(n^2)\).
    • Therefore: \(\Theta(n^2)\) algorithm.
  • Solution & Implications
    • Solution found is (unique) man-optimal solution: each man is matched with highest-ranked woman possible in any stable marriage matching.
    • So the order in which free men are considered does not affect the result.
    • Implications for traditional courtship? Strictly speaking, no, but it is suggestive.
    • Other implications: algorithm has been used to match medical-school graduates to residency positions at hospitals, with hospitals playing the role of men.