We show that Not-All-Equal 3-Sat remains NP-complete when restricted to instances that simultaneously satisfy the following properties: (i) The clauses are given as the disjoint union of k partitions, for any fixed $k \geq 4$, of the variable set into subsets of size 3, and (ii) each pair of distinct clauses shares at most one variable. Property (i) implies that each variable appears in exactly $k$ clauses and each clause consists of exactly 3 unnegated variables. Therewith, we improve upon our earlier result (Darmann and D\"ocker, 2020). Complementing the hardness result for at least $4$ partitions, we show that for $k\leq 3$ the corresponding decision problem is in P. In particular, for $k\in \{1,2\}$, all instances that satisfy Property (i) are nae-satisfiable. By the well-known correspondence between Not-All-Equal 3-Sat and hypergraph coloring, we obtain the following corollary of our results: For $k\geq 4$, Bicolorability is NP-complete for linear 3-uniform $k$-regular hypergraphs even if the edges are given as a decomposition into $k$ perfect matchings; with the same restrictions, for $k \leq 3$ Bicolorability is in P, and for $k \in \{1,2\}$ all such hypergraphs are bicolorable. Finally, we deduce from a construction in the work by Pilz (Pilz, 2019) that every instance of Positive Planar Not-All-Equal Sat with at least three distinct variables per clause is nae-satisfiable. Hence, when restricted to instances with a planar incidence graph, each of the above variants of Not-All-Equal 3-Sat turns into a trivial decision problem.

翻译：我们证明，当限制于同时满足以下性质的实例时，Not-All-Equal 3-SAT问题仍然是NP完全的：(i) 子句被表示为变量集划分为k个部分的并集（对于任意固定的$k \geq 4$），每个部分由大小为3的子集构成；(ii) 任意两个不同子句至多共享一个变量。性质(i)意味着每个变量恰好出现在$k$个子句中，且每个子句恰好包含3个非否定变量。由此，我们改进了我们先前的结果（Darmann和Döcker，2020）。作为至少$4$个划分时硬度结果的补充，我们证明对于$k\leq 3$，相应的判定问题属于P。特别地，对于$k\in \{1,2\}$，所有满足性质(i)的实例都是nae可满足的。通过Not-All-Equal 3-SAT与超图着色的著名对应关系，我们得到以下推论：对于$k\geq 4$，线性3-均匀$k$-正则超图的二色性问题是NP完全的，即使其边集被分解为$k$个完美匹配；在相同限制下，对于$k \leq 3$，二色性问题属于P，且对于$k \in \{1,2\}$，所有此类超图都是二可着色的。最后，我们从Pilz的工作（Pilz，2019）中的一个构造推导出：每个子句至少包含三个不同变量的正平面Not-All-Equal Sat实例都是nae可满足的。因此，当限制于具有平面关联图的实例时，上述Not-All-Equal 3-SAT的每个变体都转化为一个平凡的判定问题。