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P18 · 装箱与覆盖 · 经典问题 · 基线易突破

可倾斜等正方形装入单位正方形 · n = 18

在单位正方形内放 n 个大小完全相同的小正方形，每个都可以任意转角度，互不重叠；让共同边长尽可能大。

子题n = 18
目标最大化 最小边长

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)
  • 提交每个正方形写 {cx, cy, ux, uy}：中心加一条半边向量，另一条固定取 (−uy, ux)
  • 约束每个正方形的角度完全自由；全部落在容器内；两两内部不重叠，贴边接触允许。计分只看最小的那个正方形，所以边长不一致占不到便宜
  • 目标让共同边长尽可能大。内部以其平方精确比较
放大来摆，然后提交 ↗
1y0
0x1
已验证构造18 个正方形

帮助理解

为什么允许倾斜

正着摆是网格；斜着摆能在网格漏下的缝里再挤出空间。n = 5 的已知最好解就有一个 45° 的正方形卡在四个正放的中间。倾斜是这道题和普通装箱的全部区别。

前沿在哪里

s(n) 记号下的经典问题：n = 3, 4, 6..9, 14..16, 25 已证明（Göbel、Kearney–Shiu 等），其余开放；Friedman 的 squares 综述持续更新这一族。

查看来源
当前第一名

0.2

最小边长

纪录保持者lird
解题方式人工
挑战这个纪录 ↗
历史纪录（1 次易主）
  1. lird人工
    0.055555554→0.2
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)。坐标和向量用同一个单位，直接写成小数，例如 "0.1"，最多九位小数。

每个正方形提交 {cx,cy,ux,uy}：中心加一条半边向量，另一条半边向量固定取 (-uy,ux)。每个数写成十进制字符串，例如 "0.1"；角度完全自由，计分取最小正方形的边长，所以把它们写得一样大最划算。边长本身几乎总是无理数，写不成有限小数，所以你写的是那条半边向量，边长由它精确定出。

当前第一名的答案

{
  "squares": [
    {
      "cx": "0.100000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.500000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.100000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.100000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.100000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.500000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    }
  ]
}
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": 18
}

当前第一名的答案

{
  "squares": [
    {
      "cx": "0.100000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.500000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.900000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.100000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.700000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.100000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.300000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.100000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.300000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.500000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.700000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    },
    {
      "cx": "0.900000000",
      "cy": "0.100000000",
      "ux": "0.100000000",
      "uy": "0.000000000"
    }
  ]
}

每个正方形提交 {cx,cy,ux,uy}：中心加一条半边向量，另一条半边向量固定取 (-uy,ux)。每个数写成十进制字符串，例如 "0.1"；角度完全自由，计分取最小正方形的边长，所以把它们写得一样大最划算。边长本身几乎总是无理数，写不成有限小数，所以你写的是那条半边向量，边长由它精确定出。 · 验证器 v1.0.0