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Unformatted text preview: CS 70 Discrete Mathematics and Probability Theory Fall 2010 Tse/Wagner Note 4 The Stable Marriage Problem: An Application of Proof Techniques to Analysis of Algorithms 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 (ties are not allowed). 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 For instance, the preference lists above mean that woman A is man 1s top choice; woman B is his second choice; and so on. 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). This is trouble: Before long, 1 and B are going to spending many late nights doing CS70 problem sets together. Obviously, the existence of rogue couples is not a good thing if you are a matchmaker, since they will lead to instability or customer dissatisfaction. That is why we focus on stable pairings. 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. Nonetheless, it is a stable pairing, since there are no rogue couples. The problem facing us is to find a stable pairing, given the preference lists for all n men and all n women. CS 70, Fall 2010, Note 4 1 Does a Stable Pairing Always Exist? 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 2let us consider a slightly different scenario, the roommates problem....
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 Fall '11
 Rau
 Algorithms

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