In the stable marriage problem, we are given a set of men, a set of women, and each person’s preference list. Our task is to find a stable matching, that is, a matching admitting no unmatched (man, woman)-pair each of which improves the situation by being matched together. It is known that any instance admits at least one stable matching. In this paper, we consider a natural extension where k(≥2)\documentclass[12pt]{minimal}
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\begin{document}$$k (\ge 2)$$\end{document} sets of preference lists Li\documentclass[12pt]{minimal}
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\begin{document}$$L_{i}$$\end{document} (1≤i≤k\documentclass[12pt]{minimal}
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\begin{document}$$1 \le i \le k$$\end{document}) over the same set of people are given, and the aim is to find a jointly stable matching, a matching that is stable with respect to all Li\documentclass[12pt]{minimal}
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\begin{document}$$L_{i}$$\end{document}. We show that the decision problem is NP-complete for the following two restricted cases; (1) k=2\documentclass[12pt]{minimal}
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\begin{document}$$k=2$$\end{document} and each person’s preference list is of length at most four, and (2) k=4\documentclass[12pt]{minimal}
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\begin{document}$$k=4$$\end{document}, each man’s preference list is of length at most three, and each woman’s preference list is of length at most four. On the other hand, we show that it is solvable in linear time for any k if each man’s preference list is of length at most two (women’s lists can be of unbounded length). We also show that if each woman’s preference lists are same in all Li\documentclass[12pt]{minimal}
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\begin{document}$$L_{i}$$\end{document}, then the problem can be solved in linear time.
