algorithmstable-marriage

counter example of a stable matching


I am currently reading an algorithm book and came across the Stable Matching Problem. And a question came to mind that I'm curious about, but the book doesn't answer. The question is:
For any matching, if it is not stable, pick any blocking pair(w, m), and match them. And also match their previous partners. And repeat. Is this a correct algorithm to reach a stable matching?
It seems the answer is no. But I can not think out a counter example. Is there anyone who can help?


Solution

  • I think I have found the answer. Suppose we have 3 women and 3 men. The preference list of them are:
    w1: m3 > m2 > m1
    w2: m2 > m3 > m1
    w3: don't care

    m1: don't care
    m2: w1 > w2 > w3
    m3: w2 > w1 > w3

    The initial matching: (w1,m1) (w2,m2) (w3,m3)
    Step 1: w1 and m2 match, then (w1,m2) (w2,m1) (w3,m3)
    Step 2: w1 and m3 match, then (w1,m3) (w2,m1) (w3,m2)
    Step 3: w2 and m3 match, then (w2,m3) (w1,m1) (w3,m2)
    Step 4: w2 and m2 match, then (w1,m1) (w2,m2) (w3,m3)

    After 4 steps, the matching goes to the initial state, which leads to an infinite loop.