Maximum Egalitarian Cost in the Stable Marriage Problem
Assume that we have n men and n women, all of whom want to get married. They rank each other without ties. After that, we can match them into n pairs for marriage. The matching is called stable if there are no rogue couples. A rogue couple is defined as a man and a woman who prefer each other over their assigned future spouses.
A famous theorem says that, whatever all the rankings are, a stable matching always exists. But how good could a stable matching be? There is a way to assign a quality score to a matching, called the egalitarian cost. The egalitarian cost of any matching is the sum of the rankings that each person gave their assigned partner. The best potential outcome is when all people are matched with their first choices. This corresponds to the minimum egalitarian cost of 2n. But what is the maximum egalitarian cost of a stable matching? I couldn’t find it in the literature, so I proved that it is n(n+1).
Proof. It is easy to see that the egalitarian cost of n(n+1) is achievable. For example, if all men gave an identical ranking to the women, and vice versa, the matching algorithm will end up with couples having mutual rankings (j,j) for different values of j. Another example is a Latin preference profile. Each woman ranks a man n+1−x whenever he ranks her x. In this case, every potential couple’s mutual rankings sum to n+1. Thus, any matching for such rankings ends up with the egalitarian cost of n(n+1).
The next step is to prove that the egalitarian cost can’t be greater. Suppose the cost of a stable matching is C and is greater than n(n+1). Then, for every person, we count the number of people who are better (ranked smaller) than their assigned partner and sum these numbers over all the people. The result must be C−2n, which is the number of pairs of people of opposite genders such that the first person prefers the second one over their assigned partner. Moreover, the result is greater than n(n−1).
The total number of possible couples is n2. Thus, the number of unrealized potential couples is n(n−1). We can conclude that we counted one of these unrealized couples twice. In such a couple, two people prefer each other over their assigned partners. Thus, they form a rogue couple, contradicting our assumption that the matching is stable.Share:
Nice simple one! I will remember it 😀 matching firms, graduate schools, and mates.5 June 2022, 9:28 pm
I suppose the next question is weather or not we can find an algorithm to always get the Minimum (or Maximum) Egalitarian Cost Stable Matching (or just Matching)? So 4 nice problems!6 June 2022, 2:49 am