Python 中的最小封闭平行四边形
Minimal enclosing parallelogram in Python
我有一组定义凸多边形的点，我想使用 Python and/or NumPy 找到具有最小面积的封闭平行四边形。
这里有一些可能有用的资源，但我自己无法充分理解它们：
minboundparallelogram(x,y,metric) in MATLAB
Paper on a couple of proposed algorithms
非常感谢任何帮助。 O(n) 解决方案并不重要。
如果您想要最小 周长 包围平行四边形，
我下面的论文提供了一个线性时间算法。
所有最小封闭形状算法都依赖于 "rotating calipers,"
从某种意义上你可以从上图推断出来。
Bousany, Yonit, Mary Leah Karker, Joseph O'Rourke, and Leona Sparaco. "Sweeping minimum perimeter enclosing parallelograms: Optimal crumb cleanup." In Canad. Conf. Comput. Geom., pp. 167-170. 2010.
(PDF download.)
这是我使用的纯 Python O(n) 实现：
import math
"""
Minimal Enclosing Parallelogram
area, v1, v2, v3, v4 = mep(convex_polygon)
convex_polygon - array of points. Each point is a array [x, y] (1d array of 2 elements)
points should be presented in clockwise order.
the algorithm used is described in the following paper:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.53.9659&rep=rep1&type=pdf
"""
def distance(p1, p2, p):
return abs(((p2[1]-p1[1])*p[0] - (p2[0]-p1[0])*p[1] + p2[0]*p1[1] - p2[1]*p1[0]) /
math.sqrt((p2[1]-p1[1])**2 + (p2[0]-p1[0])**2))
def antipodal_pairs(convex_polygon):
l = []
n = len(convex_polygon)
p1, p2 = convex_polygon[0], convex_polygon[1]
t, d_max = None, 0
for p in range(1, n):
d = distance(p1, p2, convex_polygon[p])
if d > d_max:
t, d_max = p, d
l.append(t)
for p in range(1, n):
p1, p2 = convex_polygon[p % n], convex_polygon[(p+1) % n]
_p, _pp = convex_polygon[t % n], convex_polygon[(t+1) % n]
while distance(p1, p2, _pp) > distance(p1, p2, _p):
t = (t + 1) % n
_p, _pp = convex_polygon[t % n], convex_polygon[(t+1) % n]
l.append(t)
return l
# returns score, area, points from top-left, clockwise , favouring low area
def mep(convex_polygon):
def compute_parallelogram(convex_polygon, l, z1, z2):
def parallel_vector(a, b, c):
v0 = [c[0]-a[0], c[1]-a[1]]
v1 = [b[0]-c[0], b[1]-c[1]]
return [c[0]-v0[0]-v1[0], c[1]-v0[1]-v1[1]]
# finds intersection between lines, given 2 points on each line.
# (x1, y1), (x2, y2) on 1st line, (x3, y3), (x4, y4) on 2nd line.
def line_intersection(x1, y1, x2, y2, x3, y3, x4, y4):
px = ((x1*y2 - y1*x2)*(x3 - x4) - (x1 - x2)*(x3*y4 - y3*x4))/((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
py = ((x1*y2 - y1*x2)*(y3 - y4) - (y1 - y2)*(x3*y4 - y3*x4))/((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
return px, py
# from each antipodal point, draw a parallel vector,
# so ap1->ap2 is parallel to p1->p2
# aq1->aq2 is parallel to q1->q2
p1, p2 = convex_polygon[z1 % n], convex_polygon[(z1+1) % n]
q1, q2 = convex_polygon[z2 % n], convex_polygon[(z2+1) % n]
ap1, aq1 = convex_polygon[l[z1 % n]], convex_polygon[l[z2 % n]]
ap2, aq2 = parallel_vector(p1, p2, ap1), parallel_vector(q1, q2, aq1)
a = line_intersection(p1[0], p1[1], p2[0], p2[1], q1[0], q1[1], q2[0], q2[1])
b = line_intersection(p1[0], p1[1], p2[0], p2[1], aq1[0], aq1[1], aq2[0], aq2[1])
d = line_intersection(ap1[0], ap1[1], ap2[0], ap2[1], q1[0], q1[1], q2[0], q2[1])
c = line_intersection(ap1[0], ap1[1], ap2[0], ap2[1], aq1[0], aq1[1], aq2[0], aq2[1])
s = distance(a, b, c) * math.sqrt((b[0]-a[0])**2 + (b[1]-a[1])**2)
return s, a, b, c, d
z1, z2 = 0, 0
n = len(convex_polygon)
# for each edge, find antipodal vertice for it (step 1 in paper).
l = antipodal_pairs(convex_polygon)
so, ao, bo, co, do, z1o, z2o = 100000000000, None, None, None, None, None, None
# step 2 in paper.
for z1 in range(0, n):
if z1 >= z2:
z2 = z1 + 1
p1, p2 = convex_polygon[z1 % n], convex_polygon[(z1+1) % n]
a, b, c = convex_polygon[z2 % n], convex_polygon[(z2+1) % n], convex_polygon[l[z2 % n]]
if distance(p1, p2, a) >= distance(p1, p2, b):
continue
while distance(p1, p2, c) > distance(p1, p2, b):
z2 += 1
a, b, c = convex_polygon[z2 % n], convex_polygon[(z2+1) % n], convex_polygon[l[z2 % n]]
st, at, bt, ct, dt = compute_parallelogram(convex_polygon, l, z1, z2)
if st < so:
so, ao, bo, co, do, z1o, z2o = st, at, bt, ct, dt, z1, z2
return so, ao, bo, co, do, z1o, z2o
我有一组定义凸多边形的点，我想使用 Python and/or NumPy 找到具有最小面积的封闭平行四边形。
这里有一些可能有用的资源，但我自己无法充分理解它们：
minboundparallelogram(x,y,metric) in MATLAB
Paper on a couple of proposed algorithms
非常感谢任何帮助。 O(n) 解决方案并不重要。
如果您想要最小 周长 包围平行四边形， 我下面的论文提供了一个线性时间算法。
所有最小封闭形状算法都依赖于 "rotating calipers," 从某种意义上你可以从上图推断出来。
Bousany, Yonit, Mary Leah Karker, Joseph O'Rourke, and Leona Sparaco. "Sweeping minimum perimeter enclosing parallelograms: Optimal crumb cleanup." In Canad. Conf. Comput. Geom., pp. 167-170. 2010. (PDF download.)
这是我使用的纯 Python O(n) 实现：
import math
"""
Minimal Enclosing Parallelogram
area, v1, v2, v3, v4 = mep(convex_polygon)
convex_polygon - array of points. Each point is a array [x, y] (1d array of 2 elements)
points should be presented in clockwise order.
the algorithm used is described in the following paper:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.53.9659&rep=rep1&type=pdf
"""
def distance(p1, p2, p):
return abs(((p2[1]-p1[1])*p[0] - (p2[0]-p1[0])*p[1] + p2[0]*p1[1] - p2[1]*p1[0]) /
math.sqrt((p2[1]-p1[1])**2 + (p2[0]-p1[0])**2))
def antipodal_pairs(convex_polygon):
l = []
n = len(convex_polygon)
p1, p2 = convex_polygon[0], convex_polygon[1]
t, d_max = None, 0
for p in range(1, n):
d = distance(p1, p2, convex_polygon[p])
if d > d_max:
t, d_max = p, d
l.append(t)
for p in range(1, n):
p1, p2 = convex_polygon[p % n], convex_polygon[(p+1) % n]
_p, _pp = convex_polygon[t % n], convex_polygon[(t+1) % n]
while distance(p1, p2, _pp) > distance(p1, p2, _p):
t = (t + 1) % n
_p, _pp = convex_polygon[t % n], convex_polygon[(t+1) % n]
l.append(t)
return l
# returns score, area, points from top-left, clockwise , favouring low area
def mep(convex_polygon):
def compute_parallelogram(convex_polygon, l, z1, z2):
def parallel_vector(a, b, c):
v0 = [c[0]-a[0], c[1]-a[1]]
v1 = [b[0]-c[0], b[1]-c[1]]
return [c[0]-v0[0]-v1[0], c[1]-v0[1]-v1[1]]
# finds intersection between lines, given 2 points on each line.
# (x1, y1), (x2, y2) on 1st line, (x3, y3), (x4, y4) on 2nd line.
def line_intersection(x1, y1, x2, y2, x3, y3, x4, y4):
px = ((x1*y2 - y1*x2)*(x3 - x4) - (x1 - x2)*(x3*y4 - y3*x4))/((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
py = ((x1*y2 - y1*x2)*(y3 - y4) - (y1 - y2)*(x3*y4 - y3*x4))/((x1-x2)*(y3-y4) - (y1-y2)*(x3-x4))
return px, py
# from each antipodal point, draw a parallel vector,
# so ap1->ap2 is parallel to p1->p2
# aq1->aq2 is parallel to q1->q2
p1, p2 = convex_polygon[z1 % n], convex_polygon[(z1+1) % n]
q1, q2 = convex_polygon[z2 % n], convex_polygon[(z2+1) % n]
ap1, aq1 = convex_polygon[l[z1 % n]], convex_polygon[l[z2 % n]]
ap2, aq2 = parallel_vector(p1, p2, ap1), parallel_vector(q1, q2, aq1)
a = line_intersection(p1[0], p1[1], p2[0], p2[1], q1[0], q1[1], q2[0], q2[1])
b = line_intersection(p1[0], p1[1], p2[0], p2[1], aq1[0], aq1[1], aq2[0], aq2[1])
d = line_intersection(ap1[0], ap1[1], ap2[0], ap2[1], q1[0], q1[1], q2[0], q2[1])
c = line_intersection(ap1[0], ap1[1], ap2[0], ap2[1], aq1[0], aq1[1], aq2[0], aq2[1])
s = distance(a, b, c) * math.sqrt((b[0]-a[0])**2 + (b[1]-a[1])**2)
return s, a, b, c, d
z1, z2 = 0, 0
n = len(convex_polygon)
# for each edge, find antipodal vertice for it (step 1 in paper).
l = antipodal_pairs(convex_polygon)
so, ao, bo, co, do, z1o, z2o = 100000000000, None, None, None, None, None, None
# step 2 in paper.
for z1 in range(0, n):
if z1 >= z2:
z2 = z1 + 1
p1, p2 = convex_polygon[z1 % n], convex_polygon[(z1+1) % n]
a, b, c = convex_polygon[z2 % n], convex_polygon[(z2+1) % n], convex_polygon[l[z2 % n]]
if distance(p1, p2, a) >= distance(p1, p2, b):
continue
while distance(p1, p2, c) > distance(p1, p2, b):
z2 += 1
a, b, c = convex_polygon[z2 % n], convex_polygon[(z2+1) % n], convex_polygon[l[z2 % n]]
st, at, bt, ct, dt = compute_parallelogram(convex_polygon, l, z1, z2)
if st < so:
so, ao, bo, co, do, z1o, z2o = st, at, bt, ct, dt, z1, z2
return so, ao, bo, co, do, z1o, z2o