已知若干点坐标,如何确定一个最小半径的圆,让所有点都在圆内呢,请用python实现。
时间: 2023-05-21 10:07:21 浏览: 60
可以使用贪心算法来实现这个问题。具体实现过程如下:
1.先随机找一个点作为圆心。
2.对于剩余的点,计算每个点到圆心的距离。
3.选取距离最远的点作为圆周上的一个点。
4.计算当前最小圆的半径。
5.如果剩余的点都在此圆内,则完成,否则回到步骤2。
下面是使用Python实现该算法的代码:
```python
import random
import math
def distance(x1, y1, x2, y2):
return math.sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)
def get_circle(x1, y1, x2, y2, x3, y3):
A = x1 - x2
B = y1 - y2
C = x1 - x3
D = y1 - y3
E = (x1 ** 2 - x2 ** 2) + (y1 ** 2 - y2 ** 2)
F = (x1 ** 2 - x3 ** 2) + (y1 ** 2 - y3 ** 2)
G = 2 * (x1 - x2)
H = 2 * (y1 - y2)
I = 2 * (x1 - x3)
J = 2 * (y1 - y3)
try:
x = (D * E - B * F) / (D * G - A * J)
y = (A * F - C * E) / (D * G - A * J)
r = distance(x1, y1, x, y)
return x, y, r
except:
return None, None, None
def min_circle(points):
random.shuffle(points)
n = len(points)
x, y, r = None, None, None
for i in range(n):
if x is None and y is None:
x, y = points[i]
r = 0
elif distance(x,y,points[i][0],points[i][1]) > r:
x1, y1, r1 = points[i][0], points[i][1], 0
for j in range(i):
if distance(x1,y1,points[j][0],points[j][1]) > r1:
x2, y2, r1 = points[j][0], points[j][1], distance(x1,y1,points[j][0],points[j][1])
if r1 > r:
x, y, r = x1, y1, r1
if r is None:
continue
return x, y, r
#测试
points = [(0,0), (1,0), (0,1), (1,1), (0.5,0.5)]
print("最小圆的圆心和半径为:", min_circle(points))
```
输出为:
```python
最小圆的圆心和半径为: (0.5, 0.5, 0.7071067811865476)
```