一个更简单的Not-All-Equal 3-SAT困难变体
An even simpler hard variant of Not-All-Equal 3-SAT
AI总结：
该研究证明满足变量集k≥4个大小为3划分且子句共享变量≤1的Not-All-Equal 3-Sat仍NP完全，k≤3时属P类，还关联超图二色性及平面实例的平凡性，改进了之前结果。
摘要：
我们证明，当Not-All-Equal 3-Sat实例同时满足以下性质时，其仍为NP完全问题：(i)子句由变量集的k个划分（对于任意固定的k≥4）的不相交并组成，每个划分的子集大小为3；(ii)每对不同子句最多共享一个变量。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，我们改进了之前的结果（Darmann和Döcker，2020）。作为至少4个划分的困难性结果的补充，我们证明当k≤3时，相应的决策问题属于P类。特别地，当k∈{1,2}时，所有满足性质(i)的实例都是nae可满足的。通过Not-All-Equal 3-Sat与超图着色之间的已知对应关系，我们得出以下推论：当k≥4时，线性3一致k正则超图的二色性问题是NP完全的，即使边被分解为k个完美匹配；在相同限制下，当k≤3时二色性问题属于P类，且当k∈{1,2}时所有此类超图都是可二色的。最后，我们从Pilz（2019）的构造中推导出，每个子句至少有三个不同变量的Positive Planar Not-All-Equal Sat实例都是nae可满足的。因此，当限制在具有平面关联图的实例时，上述所有Not-All-Equal 3-Sat变体都成为平凡决策问题。
英文摘要：
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 Döcker, 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.
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