如何在矩形周围均匀分布点

I am going to suggest an algorithm, but since the derivation is a bit mathematically messy, I did not have enough time to think it through carefully and to check it carefully for correctness. Plus I might have included some redundant checks, which if one proves some proper inequalities may become redundant and one may prove the existence of a solution may always exist under reasonable assumptions. I believe the idea is correct, but I might have made some mistakes writing this thing up, so be careful.

Because according to your comment, having only one corner inside the segment along the boundary of the square is enough to solve the rest of the cases due to symmetry, I will focus on the one corner case.

Your polygonal segment with one 90 degree corner is divided into a pair of perpendicular straight line segments, the first one of length l1 and the second of length l2. These two lengths are given to you. You also want to add on the polygonal segment, which is of total length l1 + l2, a given number of n points so that the euclidean straight line distance between any two consecutive points is the same. Call that unknown distance d. When you do that you are going to end up with n1 full segments of unknown length d on l1 and n2 full segments of unknown length d on l2 so that

n1 + n2 = n

In general, you will also end up with an extra segment of length d1 <= d on l1 with one end at the 90 degree corner. Analogously, you will also have an extra segment of length d2 <= d on l2 with one end at the 90 degree corner. Thus, the two segments d1 and d2 share a common end and are perpendicular, so they form a right-angled triangle. According to Pythagoras' theorem, these two segments satisfy the equation

d1^2 + d2^2 = d^2

If we combine all the equations and information up to now, we obtain a system of equations and restrictions which are:

n1*d + d1 = l1
n2*d + d2 = l2
d1^2 + d2^2 = d^2
n1 + n2 = n
n1 and n2 are non-negative integers

where the variables d, d1, d2, n1, n2 are unknown while l1, l2, n are given. From the first two equations, you can express d1 and d2 and substitute in the third equation:

d1 = l1 - n1*d
d2 = l2 - n2*d
(l1 - n1*d)^2 + (l2 - n2*d)^2 = d^2
n1 + n2 = n
n1 and n2 are non-negative integers

In the special case, when one wants to add only one point, i.e. n = 1, one has either n1 = n = 1 or n2 = n = 1 depending on whether l1 > l2 or l1 <= l2 respectively. Say l1 > l2. Then n1 = n = 1 and n2 = 0 so

d1 = l1 - d
d2 = l2
(l1 - d)^2 + l2^2 = d^2

Expand the equation, simplify it and solve for d:

l1^2 - 2*l1*d + d^2 + l2^2 = d^2
l1^2 + l2^2 - 2*l1*d = 0
d = (l1^2 + l2^2) / (2*l1)

Next, let us go back to the general case. You have to solve the system

(l1 - n1*d)^2 + (l2 - n2*d)^2 = d^2
n1 + n2 = n
n1 and n2 are non-negative integers

where the variables d, n1, n2 are unknown while l1, l2, n are given. Expand the first equation:

l1^2  -  2 * l1 * n1 * d  +  n1^2 * d^2  +  l2^2  -  2 * l2 * n2 * d  +  n2^2 * d^2 = d^2
n1 + n2 = n
n1 and n2 are non-negative integers

and group the terms together

(n1^2 + n2^2 - 1) * d^2  - 2 * (l1*n1 + l2*n2) * d  +  (l1^2 + l2^2) = 0
n1 + n2 = n
n1 and n2 are non-negative integers

The first equation is a quadratic equation in d

(n1^2 + n2^2 - 1) * d^2  - 2 * (l1*n1 + l2*n2) * d  +  (l1^2 + l2^2) = 0

By the quadratic formula you expect two solutions (in general, you choose whichever is positive. If both are positive and d < l1 and d < l2, you may have two solutions):

d = ( (l1*n1 + l2*n2) +- sqrt( (l1*n1 + l2*n2)^2 - (l1^2 + l2^2)*(n1^2 + n2^2 - 1) ) ) / (n1^2 + n2^2 - 1)
n1 + n2 = n
n1 and n2 are non-negative integers

现在，如果您可以找到合适的n1和n2，则可以d使用上面的二次公式来计算必要的值。为了存在解，平方根中的表达式必须为非负，因此您有不等式约束

d = ( (l1*n1 + l2*n2) +- sqrt( (l1*n1 + l2*n2)^2 - (l1^2 + l2^2)*(n1^2 + n2^2 - 1) ) ) / (n1^2 + n2^2 - 1)
(l1*n1 + l2*n2)^2 - (l1^2 + l2^2)*(n1^2 + n2^2 - 1) >= 0
n1 + n2 = n
n1 and n2 are non-negative integers

简化不平等表达：

(l1*n1 + l2*n2)^2 - (l1^2 + l2^2)*(n1^2 + n2^2 - 1) = (l1^2 + l2^2) - (l1*n2 - l2*n1)^2

这将我们带入以下系统

d = ( (l1*n1 + l2*n2) +- sqrt( (l1^2 + l2^2) - (l1*n2 - l2*n1)^2 ) ) / (n1^2 + n2^2 - 1)
(l1^2 + l2^2) - (l1*n2 - l2*n1)^2 >= 0
n1 + n2 = n
n1 and n2 are non-negative integers

分解不平等因素，

d = ( (l1*n1 + l2*n2) +- sqrt( (l1^2 + l2^2) - (l1*n2 - l2*n1)^2 ) ) / (n1^2 + n2^2 - 1)
(sqrt(l1^2 + l2^2) - l1*n2 + l2*n1) * (sqrt(l1^2 + l2^2) + l1*n2 - l2*n1) >= 0
n1 + n2 = n
n1 and n2 are non-negative integers

因此，对于这些限制，您有两种情况：

情况1：

d = ( (l1*n1 + l2*n2) +- sqrt( (l1^2 + l2^2) - (l1*n2 - l2*n1)^2 ) ) / (n1^2 + n2^2 - 1)
sqrt(l1^2 + l2^2) - l1*n2 + l2*n1 >= 0 
sqrt(l1^2 + l2^2) + l1*n2 - l2*n1 >= 0
n1 + n2 = n
n1 and n2 are positive integers

或情况2：

d = ( (l1*n1 + l2*n2) +- sqrt( (l1^2 + l2^2) - (l1*n2 - l2*n1)^2 ) ) / (n1^2 + n2^2 - 1)
sqrt(l1^2 + l2^2) - l1*n2 + l2*n1 <= 0 
sqrt(l1^2 + l2^2) + l1*n2 - l2*n1 <= 0
n1 + n2 = n
n1 and n2 are positive integers

我们关注案例1，发现案例2是不可能的。从表达开始n2 = n - n1，然后将其替换为两个不等式中的每一个，并n1在每个不等式的一侧进行隔离。此过程将产生：

情况1：

d = ( (l1*n1 + l2*n2) +- sqrt( (l1^2 + l2^2) - (l1*n2 - l2*n1)^2 ) ) / (n1^2 + n2^2 - 1)
( l1*n - sqrt(l1^2 + l2^2) ) / (l1 + l2) <=  n1  <= ( l1*n + sqrt(l1^2 + l2^2) ) / (l1 + l2) 
n1 + n2 = n
n1 and n2 are positive integers

可以看到情况2逆转了不等式，这是不可能的，因为左侧小于右侧。

所以算法可能是这样的：

function d = find_d(l1, l2, n)
{
   if n = 1 and l1 > l2 { 
      return d = (l1^2 + l2^2) / (2*l1)
   } else if n = 1 and l1 <= l2 {
      return d = (l1^2 + l2^2) / (2*l2)
   }
   for integer n1 >= 0 starting from floor( ( l1*n - sqrt(l1^2 + l2^2) ) / (l1 + l2) ) to floor( ( l1*n + sqrt(l1^2 + l2^2) ) / (l1 + l2) ) + 1 
   {
      n2 = n - n1
      D = (l1^2 + l2^2) - (l1*n2 - l2*n1)^2
      if D >= 0
      {
         d1 = ( (l1*n1 + l2*n2) - sqrt( D ) ) / (n1^2 + n2^2 - 1)  
         d2 = ( (l1*n1 + l2*n2) + sqrt( D ) ) / (n1^2 + n2^2 - 1) 
         if 0 < d1 < max(l1, l2) {       
            return d1
         } else if  0 < d2 < max(l1, l2) {
            return d2
         } else {
            return "could not find a solution"
         }
      }
   }
}