A set family ${\cal F}$ is $uncrossable$ if $A \cap B,A \cup B \in {\cal F}$ or $A \setminus B,B \setminus A \in {\cal F}$ for any $A,B \in {\cal F}$. A classic result of Williamson, Goemans, Mihail, and Vazirani [STOC 1993:708-717] states that the problem of covering an uncrossable set family by a min-cost edge set admits approximation ratio $2$, by a primal-dual algorithm. They asked whether this result extends to a larger class of set families and combinatorial optimization problems. We define a new class of $semi$-$uncrossable$ $set$ $families$, when for any $A,B \in {\cal F}$ we have that $A \cap B \in {\cal F}$ and one of $A \cup B,A \setminus B ,B \setminus A$ is in ${\cal F}$, or $A \setminus B,B \setminus A \in {\cal F}$. We will show that the Williamson et al. algorithm extends to this new class of families and identify several ``non-uncrossable'' algorithmic problems that belong to this class. In particular, we will show that the union of an uncrossable family and a monotone family, or of an uncrossable family that has the disjointness property and a proper family, is a semi-uncrossable family, that in general is not uncrossable. For example, our result implies approximation ratio $2$ for the problem of finding a min-cost subgraph $H$ such that $H$ contains a Steiner forest and every connected component of $H$ contains at least $k$ nodes from a given set $T$ of terminals.

翻译：集合族${\cal F}$被称为$不可交叉的$若对于任意$A,B \in {\cal F}$，有$A \cap B, A \cup B \in {\cal F}$或$A \setminus B, B \setminus A \in {\cal F}$。Williamson、Goemans、Mihail和Vazirani的经典结果[STOC 1993:708-717]指出，通过原始-对偶算法，以最小成本边集覆盖不可交叉集合族的问题可实现近似比$2$。他们询问这一结果是否能推广到更广泛的集合族类别和组合优化问题。我们定义了一类新的$半不可交叉集合族$，即对于任意$A,B \in {\cal F}$，有$A \cap B \in {\cal F}$，且$A \cup B, A \setminus B, B \setminus A$中之一属于${\cal F}$，或$A \setminus B, B \setminus A \in {\cal F}$。我们将证明Williamson等人的算法可推广至这一新类别，并确定属于该类别的若干“非不可交叉”算法问题。特别地，我们将证明一个不可交叉族与一个单调族的并集，或一个具有不相交性质的不可交叉族与一个恰当族的并集，构成一个通常并非不可交叉的半不可交叉族。例如，我们的结果意味着对于寻找最小成本子图$H$（使得$H$包含一个Steiner森林且$H$的每个连通分量至少包含来自给定终端集$T$的$k$个节点）的问题，近似比为$2$。