用互不相交的弧连接两对平面点：精确的√2长度界
Joining Two Pairs of Planar Points by Disjoint Arcs: The Sharp $\sqrt{2}$ Length Bound
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中文总结 AI 辅助
针对平面点对连接的问题，证明了两对时互不相交弧的最小最坏长度为√2，并指出该书提出的三对的对应数值不成立。
中文摘要 AI 辅助
克罗夫特、法尔科纳和盖伊的问题F16要求，当每对点之间的距离最多为1时，用两两互不相交的平面弧连接指定点对所需的最坏情况下的最小长度。该书指出两对的答案为√2，我们对此进行了严格证明。下界是基于正方形的交叉论证，上界则来自一个精确椭圆引理。由此可得，该书提出的三对的数值(√3+1)/2在字面表述下不可能正确，因为该数值随点对数量增加而非递减。
英文摘要
Problem F16 of Croft, Falconer, and Guy asks for the least worst-case length needed to join prescribed pairs of points by pairwise disjoint planar arcs, when the distance within each pair is at most one. The book suggests that the answer for two pairs is $\sqrt2$. We prove this exactly. The lower bound is a crossing argument in a square. The upper bound follows from a sharp ellipse lemma. As a consequence, the value proposed in the book for three pairs, $(\sqrt3+1)/2$, cannot be correct under the literal formulation, because the constants are nondecreasing in the number of pairs.
