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algorithm.design.kleinberg.tardos.solutions.ch1 (5)

# algorithm.design.kleinberg.tardos.solutions.ch1 (5) - (a...

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Unformatted text preview: (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 7 that is if a man m is indifferent between two women w,- and wj then 10,- appears on m’s preference list before wj ifi < j and ifj < i wj appears before 10,. Similarly if to is indifferent between two men m,- and m, then m,- appears on Ms preference list before 771, ifi < j and ifj < i m, appears before m,. Now that we have concrete preference lists, we run the stable matching algorithm. We claim that the matching produced would have no strong instability. But this 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 7 a matching with no instabilities. (b) The answer is No. The following is a simple counterexample. Let n : 2 and m1, 7112 be the two men, and 101,102 the two women. Let m1 be indifferent between wl and 102 and let both of the women prefer ml to 7712. The choices of 7112 are insigniﬁcant. There is no matching without weak stability in this example, since regardless of who was matched with ml, the other woman together with m1 would form a weak instability. 1ex734.923.393 ...
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