# aa1-sol - b 1 g 1 = 1 g 2 b 2 it can be anything g 1 b 1...

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Problem Set 1 Solutions Due: Feb 2 cs2me3 Winter 2007 Part B 1. (a) The answer is Yes. A simple way to think about it is to break the ties in some fashion and then run the stable matching algorithm on the resulting preference lists. We can for example break the ties lexicographically—that is, if a b is indiﬀerent between two g i , g j then g i appears on b ’s preference list before g j if i < j and if j < i g j appears before g i . Similarly, if g is indiﬀerent between b i , b j then b i appears on g ’s preference list before b j if i < j and if j < i b j appears before b i . Now that we have a set of concrete preference lists, we run the stable matching algorithm. We claim that the matching produced would have no strong instability. But the latter claim is true because any strong instability would be an instability for the match produced by the algorithm, yet we know that the algorithm produced a stable matching—a matching with no instabilities. (b) The answer is No. The following is a simple counterexample. Let n = 2, and consider the following preference lists:
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Unformatted text preview: b 1 g 1 = 1 g 2 b 2 it can be anything g 1 b 1 < 1 b 2 g 2 b 1 < 2 b 2 There is no matching without weak instability in this example, since regardless of who was matched with b 1 , the other g together with b 1 would form a weak instability. 2. We do (b); it is possible to get a more desirable partner by falsely switching order of the preferences. Assume that we have three b ’s and three g ’s with the preference lists given in the table below, where the last column we have a false list of preferences of g 3 . b 1 b 2 b 3 g 1 g 2 g 3 g 3 g 3 g 1 g 3 b 1 b 1 b 2 b 2 g 1 g 2 g 1 b 2 b 2 b 1 b 3 g 2 g 2 g 2 b 3 b 3 b 3 b 1 Now run the algorithm with the true column, and note that we obtain the following pairing: ( b 1 , g 3 ) , ( b 2 , g 1 ) , ( b 3 , g 2 ). On the other hand, if we run the algorithm with the false column we get the pairing ( b 1 , g 1 ) , ( b 2 , g 3 ) , ( b 3 , g 2 ). So g 3 ends up with m 2 who is her true favorite. 1...
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## This note was uploaded on 02/10/2009 for the course RLG 100 taught by Professor Pearly during the Spring '08 term at University of Toronto.

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