In the Stable Matching Problem, it is stated that there can exist cases where th

Question

In the Stable Matching Problem, it is stated that there can exist cases where the m list of men can be content with their decisions, yet the list of f cannot when the algorithm is run with men's proposals.

From what I read, an unstable match occurs when m and f prefer each other to their current partners.

I am a little lost in the definition of Stable Matching for this case. I'm going over the slides here.

Is a pair (m,f) stable as long as the men are content even though the female's preferences have not been matched?

Explanation / Answer

Yes, it is stable. It doesn't need to assign the optimal choices for both sides. To break a marriage you need two willing parties, unhappiness of one side in a marriage doesn't make it unstable here.

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