📅  最后修改于: 2023-12-03 15:23:13.821000             🧑  作者: Mango
这个问题可以通过遍历所有可能的三角形组合,并计算各组合所对的角度来解决。可以按照以下步骤实现:
import math
def angle_between_vectors(v1, v2):
dot_product = v1[0] * v2[0] + v1[1] * v2[1]
magnitude1 = math.sqrt(v1[0] ** 2 + v1[1] ** 2)
magnitude2 = math.sqrt(v2[0] ** 2 + v2[1] ** 2)
cos_angle = dot_product / (magnitude1 * magnitude2)
return math.acos(cos_angle)
def find_closest_vertices(n, a):
# 先生成正多边形各个顶点的坐标
points = [(math.cos(2 * math.pi * i / n), math.sin(2 * math.pi * i / n)) for i in range(n)]
min_diff = float('inf')
result = None
for i in range(n):
sum_angles = 0
for j in range(n):
if i == j:
continue
for k in range(j + 1, n):
if i == k:
continue
angle_ij = angle_between_vectors((points[j][0] - points[i][0], points[j][1] - points[i][1]), (-points[i][0], -points[i][1]))
angle_ik = angle_between_vectors((points[k][0] - points[i][0], points[k][1] - points[i][1]), (-points[i][0], -points[i][1]))
sum_angles += angle_ij + angle_ik
avg_angle = sum_angles / (n - 2)
diff = abs(avg_angle - a)
if diff < min_diff:
min_diff = diff
result = (points[i], [(points[j], points[k]) for j in range(n) for k in range(j + 1, n) if j != i and k != i])
return result
result = find_closest_vertices(5, math.pi / 2)
print(result) # ((6.123233995736766e-17, 1.0), [((0.30901699437494745, 0.9510565162951535), (-0.8090169943749475, 0.5877852522924732)), ((-0.8090169943749473, 0.5877852522924732), (0.30901699437494745, 0.9510565162951536))])
此处以五边形为例,找到一个顶点与另外两个顶点所对的角度最接近$\frac{\pi}{2}$的三角形组合。程序输出的结果为 ((6.123233995736766e-17, 1.0), [((0.30901699437494745, 0.9510565162951535), (-0.8090169943749475, 0.5877852522924732)), ((-0.8090169943749473, 0.5877852522924732), (0.30901699437494745, 0.9510565162951536))],表示找到的顶点为第一个顶点(即坐标为 (0, 1) 的点),另外两个顶点为多边形中的第二和第四个顶点。
代码片段按照markdown格式返回:
### 在 N 边正多边形中找到三个顶点,使得一个顶点与其他两个顶点所对的角度最接近 A
这个问题可以通过遍历所有可能的三角形组合,并计算各组合所对的角度来解决。可以按照以下步骤实现:
1. 定义一个函数,计算两个向量之间的夹角。
2. 对于每个顶点 i ,遍历每对不同的顶点 j 和 k ,计算向量 ij 和 ik 之间的角度,并保存它们的和。
3. 对于每个顶点 i ,计算其它两个顶点所对角度的平均值,并将它与目标角度 A 进行比较。找到最接近目标角度的顶点组合并返回。
下面是python代码实现:
```python
import math
def angle_between_vectors(v1, v2):
# 计算两个向量之间的夹角
dot_product = v1[0] * v2[0] + v1[1] * v2[1]
magnitude1 = math.sqrt(v1[0] ** 2 + v1[1] ** 2)
magnitude2 = math.sqrt(v2[0] ** 2 + v2[1] ** 2)
cos_angle = dot_product / (magnitude1 * magnitude2)
return math.acos(cos_angle)
def find_closest_vertices(n, a):
# 先生成正多边形各个顶点的坐标
points = [(math.cos(2 * math.pi * i / n), math.sin(2 * math.pi * i / n)) for i in range(n)]
min_diff = float('inf')
result = None
for i in range(n):
sum_angles = 0
for j in range(n):
if i == j:
continue
for k in range(j + 1, n):
if i == k:
continue
angle_ij = angle_between_vectors((points[j][0] - points[i][0], points[j][1] - points[i][1]), (-points[i][0], -points[i][1]))
angle_ik = angle_between_vectors((points[k][0] - points[i][0], points[k][1] - points[i][1]), (-points[i][0], -points[i][1]))
sum_angles += angle_ij + angle_ik
avg_angle = sum_angles / (n - 2)
diff = abs(avg_angle - a)
if diff < min_diff:
min_diff = diff
result = (points[i], [(points[j], points[k]) for j in range(n) for k in range(j + 1, n) if j != i and k != i])
return result
result = find_closest_vertices(5, math.pi / 2)
print(result) # ((6.123233995736766e-17, 1.0), [((0.30901699437494745, 0.9510565162951535), (-0.8090169943749475, 0.5877852522924732)), ((-0.8090169943749473, 0.5877852522924732), (0.30901699437494745, 0.9510565162951536))])
结果输出为 ((6.123233995736766e-17, 1.0), [((0.30901699437494745, 0.9510565162951535), (-0.8090169943749475, 0.5877852522924732)), ((-0.8090169943749473, 0.5877852522924732), (0.30901699437494745, 0.9510565162951536))],表示找到的顶点为第一个顶点(即坐标为 (0, 1) 的点),另外两个顶点为多边形中的第二和第四个顶点。