找出平面上的4个点是否形成一个矩形? [英] find if 4 points on a plane form a rectangle?

问题描述

有人可以用 C 风格的伪代码告诉我如何编写一个函数(表示你喜欢的点)，如果 4 点(函数的参数)形成一个矩形，则返回 true，否则返回 false?

我想出了一个解决方案，首先尝试找到 2 对具有相同 x 值的不同点，然后对 y 轴执行此操作.但是代码比较长.只是想看看其他人的想法.

解决方案

• 求角点的质心:cx=(x1+x2+x3+x4)/4, cy=(y1+y2+y3+y4)/4
• 测试从质心到所有 4 个角的距离的平方是否相等

<块引用>

bool isRectangle(double x1, double y1,双 x2，双 y2，双 x3，双 y3，双 x4，双 y4){双 cx,cy;双 dd1,dd2,dd3,dd4;cx=(x1+x2+x3+x4)/4；cy=(y1+y2+y3+y4)/4；dd1=sqr(cx-x1)+sqr(cy-y1);dd2=sqr(cx-x2)+sqr(cy-y2);dd3=sqr(cx-x3)+sqr(cy-y3);dd4=sqr(cx-x4)+sqr(cy-y4);返回 dd1==dd2 &&dd1==dd3 &&dd1==dd4;}

(当然在实践中测试两个浮点数 a 和 b 的相等性应该以有限的精度进行:例如 abs(a-b) < 1E-6)

Can somebody please show me in C-style pseudocode how to write a function (represent the points however you like) that returns true if 4-points (args to the function) form a rectangle, and false otherwise?

I came up with a solution that first tries to find 2 distinct pairs of points with equal x-value, then does this for the y-axis. But the code is rather long. Just curious to see what others come up with.

解决方案

• find the center of mass of corner points: cx=(x1+x2+x3+x4)/4, cy=(y1+y2+y3+y4)/4
• test if square of distances from center of mass to all 4 corners are equal

bool isRectangle(double x1, double y1,
                 double x2, double y2,
                 double x3, double y3,
                 double x4, double y4)
{
  double cx,cy;
  double dd1,dd2,dd3,dd4;

  cx=(x1+x2+x3+x4)/4;
  cy=(y1+y2+y3+y4)/4;

  dd1=sqr(cx-x1)+sqr(cy-y1);
  dd2=sqr(cx-x2)+sqr(cy-y2);
  dd3=sqr(cx-x3)+sqr(cy-y3);
  dd4=sqr(cx-x4)+sqr(cy-y4);
  return dd1==dd2 && dd1==dd3 && dd1==dd4;
}

(Of course in practice testing for equality of two floating point numbers a and b should be done with finite accuracy: e.g. abs(a-b) < 1E-6)

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