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StableMarriage - CS 70 Fall 2006 Discrete Mathematics for...

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CS 70 Discrete Mathematics for CS Fall 2006 Papadimitriou & Vazirani Lecture 6 Stable Marriage - Application of Proof Techniques for Algo- rithmic Analysis Consider a dating agency that must match up n men and n women. Each man has an ordered preference list of the n women, and each woman has a similar list of the n men. Is there a good algorithm that the agency can use to determine a good pairing? Example Consider for example n = 3 men (represented by numbers 1, 2, and 3) and 3 women ( A , B , and C ), and the following preference lists: Men Women 1 A B C 2 B A C 3 A B C Women Men A 2 1 3 B 1 2 3 C 1 2 3 What properties should a good pairing have? One possible criterion for a “good” pairing is one in which the number of first ranked choices is maximized. Another possibility is to minimize the number of last ranked choices. Or perhaps minimizing the sum of the ranks of the choices, which may be thought of as maximizing the average happiness. In this lecture we will focus on a very basic criterion: stability . A pairing is unstable if there is a man and a woman who prefer each other to their current partners. We will call such a pair a rogue couple. So a pairing of n men and n women is stable if it has no rogue couples. An unstable pairing from the above example is: { (1,C), (2,B), (3,A) } . The reason is that 1 and B form a rogue couple, since 1 would rather be with B than C (his current partner), and since B would rather be with 1 than 2 (her current partner). An example of a stable pairing is: { (2,A), (1,B), (3,C) } . Note that (1,A) is not a rogue couple. It is true that man 1 would rather be with woman A than his current partner. Unfortunately for him, she would rather be with her current partner than with him. Note also that both 3 and C are paired with their least favorite choice in this matching. Before we discuss how to find a stable pairing, let us ask a more basic question: do stable pairings always exist? Surely the answer is yes, since we could start with any pairing and make it more and more stable as follows: if there is a rogue couple, modify the current pairing so that they are together. Repeat. Surely this procedure must result in a stable pairing! Unfortunately this reasoning is not sound. To demonstrate this, let us consider a slightly different scenario, the roommates problem. Here we have 2 n people who must be paired up to be roommates (the difference being that unlike the dating scenario, a person can be paired with any of the remaining 2 n - 1). The point is that nothing about the above reasoning relied on the fact that men can only be paired with women in the dating scenario, so by the same reasoning we would expect that there would be a stable pairing for the roommates problem as well. The following counter-example illustrates the fallacy in the reasoning: Roommates Problem: CS 70, Fall 2006, Lecture 6 1
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Roommates A B C D B C A D C A B D D Visually, we have the following situation: What is interesting about this problem is that there is no stable pairing (i.e. there is always a rogue couple).
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