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The same decomposition works in higher dimensions. For NNN random points on a sphere, the probability they all lie in a hemisphere is also N/2N−1N / 2^{N-1}N/2N−1. The argument is identical: anchor a hemisphere at each point, observe that each event has probability 1/2N−11/2^{N-1}1/2N−1, and verify that at most one anchor can work (the complementary cap is at least a full hemisphere, so no other anchor's hemisphere can straddle it). TODO: If I figure out how to add 3d visualizations to this website, I'll cover the 3D case
。关于这个话题，谷歌浏览器【最新下载地址】提供了深入分析
ЦРУ поставит оружие курдским отрядам для боевых действий против Ирана08:32
关于这些隐私数据，这两天还有一篇论文在 X 上非常火，讲的其实就是老生常谈的问题，这些 AI 大模型如何使用我们的对话数据。