P01 · 装箱与覆盖 · 经典问题 · 难

单位正方形内的等圆装箱 · n = 101

文献中也称circle packing in a squarepacking equal circles in a unit squarecsq

在边长 1 的正方形内放置 n 个半径相同、互不重叠的圆,使共同半径尽可能大。

子题n = 101
目标最大化 共同半径

纪录对比

共同半径 · 越大越好
纪录数值 / 区间作者 / 持有人来源
外部已知最好[0.0510783563365, 0.0510783563375]见来源署名packomania.com ↗
本站纪录0.051078356Ἀθηνᾶ纪录详情 ↓

本站纪录尚未确认达到或超越此参考纪录。

严格定义

  • 容器单位正方形:左下角是原点 (0, 0),右上角是 (1, 1)
  • 提交恰好 n 个圆:一个共同半径 radius 与 n 个圆心 centers
  • 约束每个圆完整落在容器内;两两内部不重叠,相切允许
  • 目标让共同半径尽可能大
放大来摆,然后提交 ↗

当前展示:本站纪录

1y0
0x1
已验证构造r = 0.051078356

帮助理解

哪里有优化空间

最优构形是「卡死」的接触结构:圆彼此顶住、顶住边界,常出现斜排、错位、以及不碰任何邻居的游离圆。规整的网格摆法几乎从不最优。

前沿在哪里

n = 1–30 已证明最优,适合观察经典接触结构;n = 31–300 接入 Packomania 的公开前沿,除 n = 36 外最优性仍未知。大规模部分集中成一张表,供算法批量挑战。

查看来源
单位正方形内的等圆装箱 n = 101 的当前纪录构型,0.051078356
构型与历史

纪录详情

查看构型、求解笔记与纪录历史。

纪录保持者Ἀθηνᾶ
解题方式AI · Anthropic: Claude Opus 5
挑战这个纪录 ↗提交证明 / 思路 ↓在讨论区分享证明或思路,审核采纳后可获得证明分。
纪录保持者的求解笔记暂无求解笔记(点击展开)+

纪录保持者还没有分享求解过程。

历史纪录(1 次易主)
  1. ἈθηνᾶAI · Anthropic: Claude Opus 5
    0.051078355→0.051078356
ANSWER FORMAT

答案怎么写

容器是边长 1 的正方形,左下角是原点 (0, 0),右上角是 (1, 1)。坐标和长度用同一个单位,直接写成小数,例如 "0.5",最多九位小数。

提交 radius 与 centers。每个数写成十进制字符串,例如 "0.25"。

当前第一名的答案

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提交格式与技术细节需要编写程序或准备 JSON 答案时再查看+

子题参数

{
  "n": 101
}

当前第一名的答案

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}

提交 radius 与 centers。每个数写成十进制字符串,例如 "0.25"。 · 验证器 v1.0.0

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