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