MAX NAE-SAT is a natural optimization problem, closely related to its better-known relative MAX SAT. The approximability status of MAX NAE-SAT is almost completely understood if all clauses have the same size $k$, for some $k\ge 2$. We refer to this problem as MAX NAE-$\{k\}$-SAT. For $k=2$, it is essentially the celebrated MAX CUT problem. For $k=3$, it is related to the MAX CUT problem in graphs that can be fractionally covered by triangles. For $k\ge 4$, it is known that an approximation ratio of $1-\frac{1}{2^{k-1}}$, obtained by choosing a random assignment, is optimal, assuming $P\ne NP$. For every $k\ge 2$, an approximation ratio of at least $\frac{7}{8}$ can be obtained for MAX NAE-$\{k\}$-SAT. There was some hope, therefore, that there is also a $\frac{7}{8}$-approximation algorithm for MAX NAE-SAT, where clauses of all sizes are allowed simultaneously. Our main result is that there is no $\frac{7}{8}$-approximation algorithm for MAX NAE-SAT, assuming the unique games conjecture (UGC). In fact, even for almost satisfiable instances of MAX NAE-$\{3,5\}$-SAT (i.e., MAX NAE-SAT where all clauses have size $3$ or $5$), the best approximation ratio that can be achieved, assuming UGC, is at most $\frac{3(\sqrt{21}-4)}{2}\approx 0.8739$. Using calculus of variations, we extend the analysis of O'Donnell and Wu for MAX CUT to MAX NAE-$\{3\}$-SAT. We obtain an optimal algorithm, assuming UGC, for MAX NAE-$\{3\}$-SAT, slightly improving on previous algorithms. The approximation ratio of the new algorithm is $\approx 0.9089$. We complement our theoretical results with some experimental results. We describe an approximation algorithm for almost satisfiable instances of MAX NAE-$\{3,5\}$-SAT with a conjectured approximation ratio of 0.8728, and an approximation algorithm for almost satisfiable instances of MAX NAE-SAT with a conjectured approximation ratio of 0.8698.

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MAX NAE-SAT 之谜

MAX NAE-SAT 是一个自然优化问题，与其比较著名的相对 MAX SAT 密切相关。如果所有子句具有相同的大小 $k$，对于某些 $k\ge 2$，则 MAX NAE-SAT 的近似状态几乎完全被理解。我们将此问题称为 MAX NAE-$\{k\}$-SAT。对于 $k=2$，它本质上是著名的 MAX CUT 问题。对于 $k=3$，它与可以被三角形部分覆盖的图中的 MAX CUT 问题有关。对于$k\ge 4$，已知$1-\frac{1}{2^{k-1}}$通过选择随机分配获得的近似比是最优的，假设$P\ne NP $. 对于每$k\ge 2$，对于MAX NAE-$\{k\}$-SAT，可以获得至少$\frac{7}{8}$的近似比。因此，有一些希望对于 MAX NAE-SAT 也有一个 $\frac{7}{8}$-approximation algorithm，允许同时使用各种大小的 where 子句。我们的主要结果是 MAX NAE-SAT 没有 $\frac{7}{8}$-approximation 算法，假设独特的游戏猜想 (UGC)。事实上，即使对于 MAX NAE-$\{3,5\}$-SAT 的几乎可满足的实例（即所有子句的大小为 $3$ 或 $5$ 的 MAX NAE-SAT），也可以达到最佳近似比率，假设 UGC，最多为 $\frac{3(\sqrt{21}-4)}{2}\approx 0.8739$。使用变分法，我们将 O'Donnell 和 Wu 对 MAX CUT 的分析扩展到 MAX NAE-$\{3\}$-SAT。我们获得了一个最优算法，假设 UGC，对于 MAX NAE-$\{3\}$-SAT，对以前的算法略有改进。新算法的近似比为 $\approx 0.9089$。我们用一些实验结果补充了我们的理论结果。