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Kabale University（卡巴莱大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文研究情境分数在非平凡区间的定量性质，对NAE_3关系给出精确公式，并证明判定情境分数为零是NP完全的，同时展示有界树宽实例的精确表述与显式族的指数扩展复杂度。
AI 中文摘要
情境分数衡量在给定的局部事件概率下，可以容纳多少全局一致的概率。我们研究存在受支持的全局赋值但情境分数非平凡的量级情形。对于平衡关系提升，我们将非情境分数识别为全局同态的容量约束填充，并在有限群对称性下证明了精确的轨道容量商。对于均匀布尔关系$\NAE_3$，这变成了一个关于正常超图二染色的少数位置拥塞博弈：\\[ \NCF(e_H)=\frac{1}{3\beta(H)}. \\] 我们还获得了极小极大对偶、割多面体表述以及显式值$2/3$和$1/2$。对于精确难度，我们将图$G$映射到一个总是二可染色的锚超图$A(G)$，使得\\[ \NCF(e_{A(G)})=1 \Longleftrightarrow \chi_f(G)\le3. \\] 我们证明了关于$\chi_f(G)$的定量稳定性，并构造了一个六边纯$\NAE_3$等式小工具，在有界出现编译下保持完整非情境分数。因此，对于最大度为七的简单$3$均匀超图，即使提供了适当的二染色，判定$\CF=0$也是NP完全的。我们进一步证明\\[ \NCF(\widehat e^{\\,\vartheta}) =\vartheta+(1-\vartheta)\NCF(e). \\] 这产生了在总是可满足的固定五元开关模板上区分$\NCF=1/2$与$\NCF\ge2/3$的NP难度，以及低于$1/12$的加性难度。一个门控染色构造给出了对于每个$\varepsilon>0$在$1/2-\varepsilon$内的难度。最后，有界树宽实例允许精确的紧凑扩展表述，而显式族通过割多面体投影具有$2^{\Omega(\sqrt N)}$的扩展复杂度。
英文摘要
The contextual fraction measures how much globally consistent probability can be packed beneath prescribed local event probabilities. We study the quantitative regime in which supported global assignments exist but the contextual fraction is nontrivial. For balanced relational lifts, we identify the noncontextual fraction with a capacity-constrained packing of global homomorphisms and prove an exact orbit-capacity quotient under finite group symmetries. For the uniform Boolean relation $\NAE_3$, this becomes a minority-position congestion game on proper hypergraph two-colourings: \[ \NCF(e_H)=\frac{1}{3β(H)}. \] We also obtain a minimax dual, a cut-polytope formulation, and explicit values $2/3$ and $1/2$. For exact hardness, we map a graph $G$ to an always-two-colourable anchor hypergraph $A(G)$ with \[ \NCF(e_{A(G)})=1 \Longleftrightarrow χ_f(G)\le3. \] We prove quantitative stability in terms of $χ_f(G)$ and construct a six-edge pure-$\NAE_3$ equality gadget that preserves the full noncontextual fraction under bounded-occurrence compilation. Thus deciding $\CF=0$ is NP-complete for simple $3$-uniform hypergraphs of maximum degree seven, even with a supplied proper two-colouring. We further prove \[ \NCF(\widehat e^{\,\vartheta}) =\vartheta+(1-\vartheta)\NCF(e). \] This yields NP-hardness of distinguishing $\NCF=1/2$ from $\NCF\ge2/3$ on an always-satisfiable fixed five-ary switch template, and additive hardness below $1/12$. A gated colouring construction gives hardness within $1/2-\varepsilon$ for every $\varepsilon>0$. Finally, bounded-treewidth instances admit exact compact extended formulations, while explicit families have extension complexity $2^{Ω(\sqrt N)}$ via cut-polytope projections.
Comments20 pages, 3 Tables, 0 Figures
