CS252 Algorithms
Monday, 1 April 2024

+ Problem sets
- What's on your mind?

+ Stable matching: the problem
- the metaphor
    1962, creepiness, power of our intuitions about "engaged", etc.
- inputs & outputs
    An "instance" of this problem consists of:
        two sets M and W, same size N, finite
        preference lists
            each m in M has an ordering/ranking of the elements of W
            each w in W has an ordering/ranking of the elements of M
    output: a set of pairings S = {(m,w),...} m in M, w in W
        S is a subset of MxW
        each m and each w appears exactly once in S
        "stable"

- what is "stability"?

Instance 1
    M = {Adam, Bob}
    W = {Zelda, Yvonne}

    Adam:   Yvonne, Zelda     {(A,Y), (B,Z)} <-- stable
    Bob:    Yvonne, Zelda     {(A,Z), (B,Y)} <-- not stable
    Yvonne: Adam, Bob             A prefers Y to current partner
    Zelda:  Bob, Adam             Y prefers A to current partner
                                     so they would elope
    
Instance 2
    Adam:   Zelda, Yvonne     {(A,Y), (B,Z)} <-- stable
                                w's don't want to switch 
    Bob:    Yvonne, Zelda     {(A,Z), (B,Y)} <-- stable
                                m's don't want to switch
    Yvonne: Adam, Bob
    Zelda:  Bob, Adam



- play with examples
    - can an instance have > 1 stable matchings?
        yes (see Instance 2 above)
    - > 2?
        yes ...
    - none?
        no ... see the theorem

- applications, variations
    - The CS Match
        People could in special cases match people to 2 classes
        Don't have to rank every class
        Sizes of the sets are different
        Not everybody gets a match
        The classes have weird preference lists
            class year, must-have-to-graduate-on-time,...
    - NRMP (medical residents & hospitals)
    - Questbridge
    - College applications
    - Only one set, and assign class project partners
        "roommates problem"
    ...

+ The algorithm
- Do one by hand
    (always do a few by hand!)

    X Adam:   *Zelda, Yvonne
    X Bob:    *Yvonne, Zelda
    X Yvonne: Adam, Bob
    X Zelda:  Bob, Adam

    S = {}
        m = Bob, w = Yvonne
        w is free, engagement happens
    S = {(B,Y)}
        m = Adam, w = Zelda
        w is free, engagement happens
    S = {(B,Y), (A,Z)}
    Loop terminates

Questions
- If I have an instance with 3 stables matches, is it possible to get G-S to generate all of them (men proposing, women proposing,...?)
- What kinds of preference lists cause the slowest runtimes?
- ...

Do a N=3 example by hand

    A: *X, Y, Z
    B: *X, Z, Y
    C: Y, X, Z

    X: B, C, A
    Y: A, C, B
    Z: A, C, B

    {(B,X)}
    ...

==== WE STOPPED HERE ===


+ Correctness proof

+ Runtime complexity