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I checked some papers on two-side stable allocation/matching (marriage, worker/company, doctor/hospital), but has not found any literature on the following problem. Can someone point out if I missed some paper or help analyze the problem?

Consider two sets of nodes, $U$ and $V$. $f_u(v)$ is the non-negative (integer) preference function of $u\in U$ towards $v\in V$. Similarly, $f_v(u)$ is for $v\in V$ towards $u\in U$. Let $E\subseteq U\times V$ be a set of pairs. Each pair $(u,v)\in E$ indicates that we can never allocate one to the other.

$\pi$ is an allocation where $\pi(u)\in V$ is the node allocated to $u$ and $\pi(v)\in U$ is the node allocated to $v$. $\pi$ is weak-stable if there does not exist a pair ($u,v$) such that $f_u(v)>f_u(\pi(u))$ and $f_v(u)>f_v(\pi(v))$, which means that $u$ prefers $v$ to $\pi(u)$ and $v$ prefers $u$ to $\pi(v)$ simultaneously.

I want to compute a stable matching with maximum size, i.e. $maximize_{\pi}|\pi|$. Notice that, ties are allowed in any preference function.

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  • $\begingroup$ It seems to me that your problem is equivalent to MAX SMTI (Stable Marriage with Ties and Incomplete lists). Am I missing something? $\endgroup$ – Hiroki Yanagisawa Oct 5 '16 at 16:02
  • $\begingroup$ @HirokiYanagisawa Many thanks! I find the MAX SMTI problem in Hard Variants of Stable Marriage. They show that this problem is NP-hard and give a 2-approximation algorithm. $\endgroup$ – user2789928 Oct 5 '16 at 18:35
  • $\begingroup$ @HirokiYanagisawa I see more than 200 papers citing it, I am wondering if you know any paper solves the weighted version, i.e. instead of simply maximizing $|\pi|$, I wish to $maximize_{\pi}\sum_{(u,v)\in \pi}f_u(v)$. I guess there will also be a 2-approximation on the weighted version. $\endgroup$ – user2789928 Oct 5 '16 at 18:36
  • $\begingroup$ I notice that there is a paper Stable Marriage with Incomplete Lists and Ties. This paper works on minimizing a particular cost (or called preference index), but not the weight I wish to maximize -- the mathmatical things are different. $\endgroup$ – user2789928 Oct 5 '16 at 18:41
  • $\begingroup$ I posted my comment in the answer section. $\endgroup$ – Hiroki Yanagisawa Oct 6 '16 at 14:44
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Your problem is equivalent to MAX SMTI (Stable Marriage with Ties and Incomplete lists). You can find the current best approximation algorithm for MAX SMTI in the following paper: Z. Kiraly, Linear Time Local Approximation Algorithm for Maximum Stable Marriage, Algorithms 2013, 6, 471-484.

There are many papers on MAX SMTI, but unfortunately I am not aware of any paper that handles the weighted version of MAX SMTI.

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