The Stable Matching Problem

Objectives

1. Define and understand the stable matching problem.

Notes

• This is from chapter 1 of our book.
  • I am not really able to follow a book, but I will somewhat try.
  • We will need to use other sources as some material is not covered.
  • And some material is not in the syllabus.
• Chapter 1 gives us a nice overview of the semester however.
  • We will study an algorithm in detail, and return to it later.
  • We will look at a few other algorithms that will illustrate future concepts.
  • And I might throw in an additional one.
• Given a set of n men and n women, all wising to find a match, find this match so that no two unmatched individuals prefer each other to their current partners.
  • The input is:
    • A set of men, M, each with a ranked list of the women.
    • A set of women, W, each with a ranked list of the men.
    • Limitations:
      • |M| = |W|
      • For any preference list, no two preferences are equal.
  • The output is a set of pairs (m_i,w_j) that is stable.
• But what is stable?
  • This is more easily explained by an example.
    • Assume we have two men {Abe, Bob}, and two women {Anne, Betty}.
    • And the preferences are as follows

    • Abe Anne Betty Bob Betty Anne

      Anne Abe Bob Betty Bob Abe

    • The Set {(Abe, Betty), (Bob, Anne)} is in-stable
      • Abe prefers Anne to Betty
      • Anne prefers Abe to Bob,
      • So there is no "cost" for them to change partners.
      • Note Betty and Bob have no say in this issue, but they are happy with the swap as well.
    • The Set {(Abe, Anne), (Bob, Betty)} is stable
      • You can't find a pair that would be happier in a different situation.
    • Let's try again with a difference preference table

    • Abe Anne Betty Bob Anne Betty

      Anne Abe Bob Betty Bob Abe

    • {(Abe, Betty), (Bob, Anne)}
      • Is this stable? IE, Would (Abe, Anne) or (Bob, Betty) be a better match?
      • Bob is happy, but Anne is not, she would be happier with Abe
      • Betty is happy, but Abe is not, he would be happier with Anne
      • So (Abe, Anne) again would be a better match
      • So this is in-stable.
    • {(Abe, Anne), (Bob, Betty)} is stable.
  • here is a more complex example from An MIT OpenCourseWare Video on this topic.
    • Men are numbered 1 through 5
    • Women are numbered A through E
    • The preference list is:

    • 1 C B E A D 2 A B E C D 3 D C B A E 4 A C D B E 5 A B D E C

      A 3 5 2 1 4 B 5 2 1 4 3 C 4 3 5 1 2 D 1 2 3 4 5 E 2 3 4 1 5

  • In the video he does a greedy match on the men. Let's do one on the women.

    Start with woman A, 
       she prefers 3 
       (3,A) is a match
       S = { (3,A) } 
    Removing both from the table, we get:
              
    
              
    1
    C
    B
    E
    D
              
    2
    B
    E
    C
    D
              
    4
    C
    D
    B
    E
              
    5
    B
    D
    E
    C
              
    
              
    B
    5
    2
    1
    4
              
    C
    4
    5
    1
    2
              
    D
    1
    2
    4
    5
              
    E
    2
    4
    1
    5
              
    Next woman B prefers 5
    (5,B) is a match
    S = { (3,A), (5,B) } 
    
    Removing both from the table, we get:
              
    
              
    1
    C
    E
    D
              
    2
    E
    C
    D
              
    4
    C
    D
    E
              
    
              
    C
    4
    1
    2
              
    D
    1
    2
    4
              
    E
    2
    4
    1
              
    Next woman C prefers 4, 
    (4,C) is a match
    S = { (3,A), (5,B), (4,C) }
    
    Removing both from the table, we get:
              
    
              
    1
    E
    D
              
    2
    E
    D
              
    
              
    D
    1
    2
              
    E
    2
    1
              
    Next woman D prefers 1
    (1,D) is a match
    S = { (3,A), (5,B), (4,C), (1,D) }
    
    Removing both from the table, we get:
              
    
              
    2
    E
              
    
              
    E
    2
              
              
    Finally, E is left with 2.
    (2, E) is a match.
    S = { (3,A), (5,B), (4,C), (1,D), (2,E) }
           

  • A quick check of this solution:
    • Each woman has her first choice
    • So no woman will want to change partners.
    • so there will be no in-stable match.
  • So is this an algorithm to find the solution?
    • No the case was really unique.
      • Each woman had a different/unique "first choice".
      • So the algorithm worked in this case.
  • But if we try this algorithm from the men's perspective we could get
    • S = { (1,C), (2,A), (3, D), (4, B), (5,E)}
    • Looking at (4,B).
      • 4 would be happier with {A, C, D}
      • C would be happier with {4, 3, 5}
      • So (4, C) would be a more stable match.
    • Other solutions are possible depending on how we chose the man who selects first
  • So greedy, at least in this formulation is not a solution.
    • We have a counter example that shows it does not work!
• Before we tackle the solution, let's make some definitions
  • A matching is a set of ordered pairs S, from MxW with the property that where each m ∈M and each w ∈ W appears only once.
    • {(1, A), (2, B) } is a match.
    • { (1, A), (2, A)} is not a match.
  • A perfect match is a matching where every member of M and W appear exactly once.
    • {(1, A), (2, B), (3, C), (4, D), (5, E)} is a perfect match.
    • {(1, A), (2, B), (3, C), (4, D) } is not a perfect match.
  • A matching S is stable if
    • S is a perfect match.
    • There is no pairing (m,w) such that
      • (m,w) ∉ S but
        • (m, #) ∈ S, (α,w) ∈ S, m ≠ α, w ≠ #
        • m prefers w to # and w prefers m to α
      • IE there are no in-stable matches.
  • The stable match problem is
    • Given a set M, each with a preference list with no ties.
    • And a set W, each with a preference list with no ties.
    • |M| = |W|
    • Find a stable match S.
  • Is this a real problem?
    • Actually it is a simplified version of many problems.
    • Applicants and jobs/internships/colleges