[MV18] introduced a fundamental new algorithmic question on stable matching, namely finding a matching that is stable under two ``nearby'' instances, where ``nearby'' meant that in going from instance $A$ to $B$, only one agent changes its preference list. By first establishing a sequence of structural results on the lattices of $A$ and $B$, [MV18] and [GMRV22] settled all algorithmic questions related to this case. The current paper essentially settles the general case. Assume that instance $B$ is obtained from $A$, both on $n$ workers and $n$ firms, via changes in the preferences of $p$ workers and $q$ firms. If so, we will denote the change by $(p, q)$. Thus [MV18] and [GMRV22] settled the case $(0, 1)$, since they adopt the convention that one firm changes its preferences. Let $\mathcal{M}_A$ and $\mathcal{M}_B$ be the sets of stable matchings of instances $A$ and $B$, and let $\mathcal{L}_A$ and $\mathcal{L}_B$ be their lattices. Our results are: 1. For $(0, n)$, $\mathcal{M}_A \cap \mathcal{M}_B$ is a sublattice of $\mathcal{L}_A$ and of $\mathcal{L}_B$. We can efficiently obtain the worker-optimal and firm-optimal stable matchings in $\mathcal{M}_A \cap \mathcal{M}_B$. We also obtain the associated partial order, as promised by Birkhoff's Representation Theorem, and use it to enumerate these matchings with polynomial delay. 2. For $(1, n)$, the only missing results are the partial order and enumeration. 3. We give an example with $(2, 2)$ for which $\mathcal{M}_A \cap \mathcal{M}_B$ fails to be a sublattice of $\mathcal{L}_A$. In light of the fact that for $(n, n)$, determining if $(\mathcal{M}_A \cap \mathcal{M}_B) = \emptyset$ is NP-hard [MO19], a number of open questions arise; in particular, closing the gap between $(2, 2)$ and $(n, n)$.

翻译：[MV18]针对稳定匹配提出了一项基础性算法新问题，即寻找在两个“邻近”实例下均稳定的匹配。其中“邻近”指从实例$A$到$B$的转换中，仅有一个代理更改其偏好列表。通过首先建立关于$A$和$B$格结构的一系列结论，[MV18]与[GMRV22]解决了与此情形相关的所有算法问题。当前论文本质上解决了该问题的一般情形。假设实例$B$由$A$（两者均包含$n$名工人和$n$个企业）经$p$名工人和$q$个企业的偏好变动得到，我们将此变动记为$(p, q)$。由于[MV18]和[GMRV22]采用了一个企业改变其偏好的惯例，他们解决了$(0, 1)$情形。令$\mathcal{M}_A$和$\mathcal{M}_B$分别为实例$A$与$B$的稳定匹配集合，$\mathcal{L}_A$与$\mathcal{L}_B$为其对应格。我们的结果包括：1. 对$(0, n)$情形，$\mathcal{M}_A \cap \mathcal{M}_B$是$\mathcal{L}_A$与$\mathcal{L}_B$的子格。我们可高效求得$\mathcal{M}_A \cap \mathcal{M}_B$中的工人最优与企业最优稳定匹配，同时依据Birkhoff表示定理得到关联偏序，并利用该偏序以多项式延迟枚举这些匹配。2. 对$(1, n)$情形，仅缺失偏序与枚举结果。3. 我们给出$(2, 2)$情形的反例，其中$\mathcal{M}_A \cap \mathcal{M}_B$不再是$\mathcal{L}_A$的子格。鉴于判断$(n, n)$情形下$(\mathcal{M}_A \cap \mathcal{M}_B) = \emptyset$的NP难度[MO19]，一系列开放问题随之产生，尤其需填补$(2, 2)$与$(n, n)$之间的理论空白。