Python 如何检查给定的四个点是否形成一个正方形(How to check if given four points form a square)

本文链接：https://blog.youkuaiyun.com/hefeng_aspnet/article/details/145685981

给定平面上四个点的坐标，判断这四个点是否形成正方形。

要检查正方形，我们需要检查以下内容:
        a) 由点形成的所有四条边都相同。
        b) 任何两条边之间的角度都是 90 度。（此条件是必需的，因为菱形也有相同的边）
        c) 检查两条对角线的距离是否相同.

例子：

输入： p1 = { 20, 10 }, p2 = { 10, 20 }, p3 = { 20, 20 }, p4 = { 10, 10 }
输出：是
解释：

输入： p1 = { 20, 20 }, p2 = { 10, 20 }, p3 = { 20, 20 }, p4 = { 10, 10 }
输出：否

方法：思路是选取任意一点并计算其与其余点的距离。设选取的点为“p”。要形成正方形，两个点与“p”的距离必须相同，设该距离为 d。与一个点的距离必须不同于 d，并且必须等于 d 的 2 倍。设距离不同的这个点为“q”。

上述条件还不够好，因为距离不同的点可能在另一侧。我们还需要检查 q 是否与其他 2 个点的距离相同，并且该距离与 d 相同。

以下是上述想法的实现：

# A Python3 program to check if
# four given points form a square or not.
class Point:

# Structure of a point in 2D space
def __init__(self, x, y):
self.x = x
self.y = y

# A utility function to find square of
# distance from point 'p' to point 'q'
def distSq(p, q):
return (p.x - q.x) * (p.x - q.x) +\
(p.y - q.y) * (p.y - q.y)

# This function returns true if (p1, p2, p3, p4)
# form a square, otherwise false
def isSquare(p1, p2, p3, p4):

d2 = distSq(p1, p2) # from p1 to p2
d3 = distSq(p1, p3) # from p1 to p3
d4 = distSq(p1, p4) # from p1 to p4

if d2 == 0 or d3 == 0 or d4 == 0:
return False

# If lengths if (p1, p2) and (p1, p3) are same, then
# following conditions must be met to form a square.
# 1) Square of length of (p1, p4) is same as twice
# the square of (p1, p2)
# 2) Square of length of (p2, p3) is same
# as twice the square of (p2, p4)

if d2 == d3 and 2 * d2 == d4 and \
2 * distSq(p2, p4) == distSq(p2, p3):
return True

# The below two cases are similar to above case
if d3 == d4 and 2 * d3 == d2 and \
2 * distSq(p3, p2) == distSq(p3, p4):
return True

if d2 == d4 and 2 * d2 == d3 and \
2 * distSq(p2, p3) == distSq(p2, p4):
return True

return False

# Driver Code
if __name__=="__main__":
p1 = Point(20, 10)
p2 = Point(10, 20)
p3 = Point(20, 20)
p4 = Point(10, 10)

if isSquare(p1, p2, p3, p4):
print('Yes')
else:
print('No')

# This code is contributed by Mayank Chaudhary
# aka chaudhary_19