问题
I'm writing a basic 2D shape library in Python (primarily for manipulating SVG drawings), and I'm at a loss for how to efficiently calculate the intersection points of two ellipses.
Each ellipse is defined by the following variables (all floats):
c: center point (x, y)
hradius: "horizontal" radius
vradius: "vertical" radius
phi: rotation from coordinate system's x-axis to ellipse's horizontal axis
Ignoring when the ellipses are identical, there could be 0 through 4 intersection points (no intersection, tangent, partially overlapping, partially overlapping and internally tangent, and fully overlapping).
I've found a few potential solutions:
- SymPy geometry module - This basically just plugs the ellipse equations into SymPy's solver. I'm not sure whether this makes sense without already having the solver. (Incidentally, I would have used SymPy instead of rolling my own, but it performs horribly when dealing with crazy floats)
- How to detect if an ellipse intersects(collides with) a circle - This could probably be adapted for two ellipses, but I'm a little fuzzy on how to turn it into sensible code.
- How Ellipse to Ellipse intersection? - The library the answer references (CADEMIA) might have a good algorithm, but I can't even figure out if it's open source.
- Wikipedia: Intersecting Two Conics - I don't have enough of a grasp of linear algebra to understand this solution.
Any suggestions on how I should go about calculating the intersections? Speed (it might have to calculate a lot of intersections) and elegance are the primary criteria. Code would be fantastic, but even a good direction to go in would be helpful.
回答1:
In math, you need to express the ellipses as bivariate quadratic equations, and solve it. I found a doucument. All the calculations are in the document, but it may take a while to implement it in Python.
So another method is to approximate the ellipses as polylines, and use shapely to find the intersections, here is the code:
import numpy as np
from shapely.geometry.polygon import LinearRing
def ellipse_polyline(ellipses, n=100):
t = linspace(0, 2*np.pi, n, endpoint=False)
st = np.sin(t)
ct = np.cos(t)
result = []
for x0, y0, a, b, angle in ellipses:
angle = np.deg2rad(angle)
sa = np.sin(angle)
ca = np.cos(angle)
p = np.empty((n, 2))
p[:, 0] = x0 + a * ca * ct - b * sa * st
p[:, 1] = y0 + a * sa * ct + b * ca * st
result.append(p)
return result
def intersections(a, b):
ea = LinearRing(a)
eb = LinearRing(b)
mp = ea.intersection(eb)
x = [p.x for p in mp]
y = [p.y for p in mp]
return x, y
ellipses = [(1, 1, 2, 1, 45), (2, 0.5, 5, 1.5, -30)]
a, b = ellipse_polyline(ellipses)
x, y = intersections(a, b)
plot(x, y, "o")
plot(a[:,0], a[:,1])
plot(b[:,0], b[:,1])
and the output:
It takes about 1.5ms on my PC.
回答2:
looking at sympy I thinks it has everything you need. (I tried to provide you with example codes; unfortunately, I failed at installing sympy with gmpy2 instead of useless python built-in mathematics)
you have :
- an ellipse with rotate method, which can be intersected with other ellipses
- or an arbitrary intersection function that takes variadic number of.. what they call as 'Geometric Entities'.
from their examples, I think it is definitely possible to intersect two ellipses:
>>> from sympy import Ellipse, Point, Line, sqrt
>>> e = Ellipse(Point(0, 0), 5, 7)
...
>>> e.intersection(Ellipse(Point(1, 0), 4, 3))
[Point(0, -3*sqrt(15)/4), Point(0, 3*sqrt(15)/4)]
>>> e.intersection(Ellipse(Point(5, 0), 4, 3))
[Point(2, -3*sqrt(7)/4), Point(2, 3*sqrt(7)/4)]
>>> e.intersection(Ellipse(Point(100500, 0), 4, 3))
[]
>>> e.intersection(Ellipse(Point(0, 0), 3, 4))
[Point(-363/175, -48*sqrt(111)/175), Point(-363/175, 48*sqrt(111)/175),
Point(3, 0)]
>>> e.intersection(Ellipse(Point(-1, 0), 3, 4))
[Point(-17/5, -12/5), Point(-17/5, 12/5), Point(7/5, -12/5),
Point(7/5, 12/5)]
edit : since general ellipse (ax^2 + bx + cy^2 + dy + exy + f) is not supported in sympy,
you should build equations and transform them yourself, and use their solver to find intersection points.
回答3:
You can use the method shown here: https://math.stackexchange.com/questions/864070/how-to-determine-if-two-ellipse-have-at-least-one-intersection-point/864186#864186
First you should be able to rescale an ellipse in one direction. To do this you need to compute the coefficients of the ellipse as a conic section, rescale, and then recover the new geometric parameters of the ellipse: center, axes, angle.
Then your problem reduces to that of finding the distance from an ellipse to the origin. To solve this last problem you need some iteration. Here is a possible self contained implementation...
from math import *
def bisect(f,t_0,t_1,err=0.0001,max_iter=100):
iter = 0
ft_0 = f(t_0)
ft_1 = f(t_1)
assert ft_0*ft_1 <= 0.0
while True:
t = 0.5*(t_0+t_1)
ft = f(t)
if iter>=max_iter or ft<err:
return t
if ft * ft_0 <= 0.0:
t_1 = t
ft_1 = ft
else:
t_0 = t
ft_0 = ft
iter += 1
class Ellipse(object):
def __init__(self,center,radius,angle=0.0):
assert len(center) == 2
assert len(radius) == 2
self.center = center
self.radius = radius
self.angle = angle
def distance_from_origin(self):
"""
Ellipse equation:
(x-center_x)^2/radius_x^2 + (y-center_y)^2/radius_y^2 = 1
x = center_x + radius_x * cos(t)
y = center_y + radius_y * sin(t)
"""
center = self.center
radius = self.radius
# rotate ellipse of -angle to become axis aligned
c,s = cos(self.angle),sin(self.angle)
center = (c * center[0] + s * center[1],
-s* center[0] + c * center[1])
f = lambda t: (radius[1]*(center[1] + radius[1]*sin(t))*cos(t) -
radius[0]*(center[0] + radius[0]*cos(t))*sin(t))
if center[0] > 0.0:
if center[1] > 0.0:
t_0, t_1 = -pi, -pi/2
else:
t_0, t_1 = pi/2, pi
else:
if center[1] > 0.0:
t_0, t_1 = -pi/2, 0
else:
t_0, t_1 = 0, pi/2
t = bisect(f,t_0,t_1)
x = center[0] + radius[0]*cos(t)
y = center[1] + radius[1]*sin(t)
return sqrt(x**2 + y**2)
print Ellipse((1.0,-1.0),(2.0,0.5)).distance_from_origin()
来源：https://stackoverflow.com/questions/15445546/finding-intersection-points-of-two-ellipses-python