解决一个三次多项式系统(以找到Bezier曲线的交点)

4
有人能建议一种修复方法或者另一种寻找该系统解决方案的路线吗? 特别是我只关心[0,1]x[0,1] 中的解(s,t)。
注意:我在这里寻找两个三次贝塞尔曲线的交点。我需要确保该方法能够找到所有的解,并且最好在合理的时间内完成(对于我的用途,这意味着每对曲线几秒钟)。
我尝试了使用sympy,但solve()和solve_poly_system()都返回了空列表。
这是我的代码:
from sympy.solvers import solve_poly_system, solve
from sympy.abc import s,t

#here are two cubics.  I'm looking for their intersection in [0,1]x[0,1]:
cub1 = 600*s**3 - 1037*s**2 + 274*s + 1237*t**3 - 2177*t**2 + 642*t + 77
cub2 = -534*s**3 + 582*s**2 + 437*s + 740*t**3 - 1817*t**2 + 1414*t - 548

#I know such a solution exists (from plotting these curves) and fsolve finds an     approximation of it no problem:
from scipy.optimize import fsolve
fcub1 = lambda (s,t): 600*s**3 - 1037*s**2 + 274*s + 1237*t**3 - 2177*t**2 + 642*t + 77
fcub2 = lambda (s,t):-534*s**3 + 582*s**2 + 437*s + 740*t**3 - 1817*t**2 + 1414*t - 548
F = lambda x: [fcub1(x),fcub2(x)]
print 'fsolve gives (s,t) = ' + str(fsolve(F,(0.5,0.5)))
print 'F(s,t) = ' + str(F(fsolve(F,(0.5,0.5))))

#solve returns an empty list
print solve([cub1,cub2])

#solve_poly_system returns a DomainError: can't compute a Groebner basis over RR
print solve_poly_system([cub1,cub2])

这将输出:
fsolve gives (s,t) = [ 0.35114023  0.50444115]
F(s,t) = [4.5474735088646412e-13, 0.0]
[]
[]

感谢阅读!


我认为这个问题不能通过sympy解决。异常是因为您在方程中使用了浮点值。在solve_poly_system的源代码中,有一行检查顺序:if a <= 2 and b <= 2 and c <= 2 and d <= 2: - HYRY
@HYRY:哦,你说的DomainError是对的。我本来想把所有小数都去掉来检查,但是我漏掉了一个!另外,你是说你认为solve_poly_system只能用于每个多项式次数小于3的系统吗? - mathandy
3个回答

4
对于贝塞尔曲线的交点,有更好的方法。(http://pomax.github.io/bezierinfo/#curveintersectionhttp://www.tsplines.com/technology/edu/CurveIntersection.pdf)。
我推荐一个简单的解决方案:实现贝塞尔细分算法(http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/Bezier/bezier-sub.html)。对于两个曲线,计算控制顶点的边界框。如果它们重叠,则可能存在交点,将其细分并使用一半重新开始这个过程(这次会有四个比较)。继续递归执行。
你不必担心指数爆炸(1、4、16、256... 次比较),因为很快许多边界框将停止重叠。
请注意,在理论上,您可以使用控制点的凸包进行计算,但在实践中,一个简单的边界框就足够了,并且更易于处理。

我认为这对我的目的很有效。我今晚会尝试实施并查看。谢谢。 - mathandy
很高兴为您服务。贝塞尔曲线具有良好的特性。 - user1196549

1

Yves的解决方案很好。这是我的代码,如果有帮助的话:

from math import sqrt
def cubicCurve(P,t):
    return P[0]*(1-t)**3 + 3*P[1]*t*(1-t)**2 + 3*P[2]*(1-t)*t**2 + P[3]*t**3
def cubicMinMax_x(points):
    local_extremizers = [0,1]
    a = [p.real for p in points]
    delta = a[1]**2 - (a[0] + a[1]*a[2] + a[2]**2 + (a[0] - a[1])*a[3])
    if delta>=0:
        sqdelta = sqrt(delta)/(a[0] - 3*a[1] + 3*a[2] - a[3])
        tau = a[0] - 2*a[1] + a[2]
        r1 = tau+sqdelta
        r2 = tau-sqdelta
        if 0<r1<1:
            local_extremizers.append(r1)
        if 0<r2<1:
            local_extremizers.append(r2)
    localExtrema = [cubicCurve(a,t) for t in local_extremizers]
    return min(localExtrema),max(localExtrema)
def cubicMinMax_y(points):
    return cubicMinMax_x([-1j*p for p in points])
def intervalIntersectionWidth(a,b,c,d): #returns width of the intersection of intervals [a,b] and [c,d]  (thinking of these as intervals on the real number line)
    return max(0, min(b, d) - max(a, c))
def cubicBoundingBoxesIntersect(cubs):#INPUT: 2-tuple of cubics (given bu control points) #OUTPUT: boolean
    x1min,x1max = cubicMinMax_x(cubs[0])
    y1min,y1max = cubicMinMax_y(cubs[0])
    x2min,x2max = cubicMinMax_x(cubs[1])
    y2min,y2max = cubicMinMax_y(cubs[1])
    if intervalIntersectionWidth(x1min,x1max,x2min,x2max) and intervalIntersectionWidth(y1min,y1max,y2min,y2max):
        return True
    else:
        return False
def cubicBoundingBoxArea(cub_points):#INPUT: 2-tuple of cubics (given bu control points) #OUTPUT: boolean
    xmin,xmax = cubicMinMax_x(cub_points)
    ymin,ymax = cubicMinMax_y(cub_points)
    return (xmax-xmin)*(ymax-ymin)
def halveCubic(P):
    return ([P[0], (P[0]+P[1])/2, (P[0]+2*P[1]+P[2])/4, (P[0]+3*P[1]+3*P[2]+P[3])/8],[(P[0]+3*P[1]+3*P[2]+P[3])/8,(P[1]+2*P[2]+P[3])/4,(P[2]+P[3])/2,P[3]])
class Pair(object):
    def __init__(self,cub1,cub2,t1,t2):
        self.cub1 = cub1
        self.cub2 = cub2
        self.t1 = t1 #the t value to get the mid point of this curve from cub1
        self.t2 = t2 #the t value to get the mid point of this curve from cub2
def cubicXcubicIntersections(cubs):
#INPUT: a tuple cubs=([P0,P1,P2,P3], [Q0,Q1,Q2,Q3]) defining the two cubic to check for intersections between.  See cubicCurve fcn for definition of P0,...,P3
#OUTPUT: a list of tuples (t,s) in [0,1]x[0,1] such that cubicCurve(cubs[0],t) - cubicCurve(cubs[1],s) < Tol_deC
#Note: This will return exactly one such tuple for each intersection (assuming Tol_deC is small enough)
    Tol_deC = 1 ##### This should be set based on your accuracy needs.  Making it smaller will have relatively little effect on performance.  Mine is set to 1 because this is the area of a pixel in my setup and so the curve (drawn by hand/mouse) is only accurate up to a pixel at most. 
    maxIts =  100 ##### This should be something like maxIts = 1-log(Tol_deC/length)/log(2), where length is the length of the longer of the two cubics, but I'm not actually sure how close to being parameterized by arclength these curves are... so I guess I'll leave that as an exercise for the interested reader :)
    pair_list = [Pair(cubs[0],cubs[1],0.5,0.5)]
    intersection_list = []
    k=0
    while pair_list != []:
        newPairs = []
        delta = 0.5**(k+2)
        for pair in pair_list:
            if cubicBoundingBoxesIntersect((pair.cub1,pair.cub2)):
                if cubicBoundingBoxArea(pair.cub1)<Tol_deC and cubicBoundingBoxArea(pair.cub2)<Tol_deC:
                    intersection_list.append((pair.t1,pair.t2)) #this is the point in the middle of the pair
                    for otherPair in pair_list:
                        if pair.cub1==otherPair.cub1 or pair.cub2==otherPair.cub2 or pair.cub1==otherPair.cub2 or pair.cub2==otherPair.cub1:
                            pair_list.remove(otherPair) #this is just an ad-hoc fix to keep it from repeating intersection points
                else:
                    (c11,c12) = halveCubic(pair.cub1)
                    (t11,t12) = (pair.t1-delta,pair.t1+delta)
                    (c21,c22) = halveCubic(pair.cub2)
                    (t21,t22) = (pair.t2-delta,pair.t2+delta)
                    newPairs += [Pair(c11,c21,t11,t21), Pair(c11,c22,t11,t22), Pair(c12,c21,t12,t21), Pair(c12,c22,t12,t22)]
        pair_list = newPairs
        k += 1
        if k > maxIts:
            raise Exception ("cubicXcubicIntersections has reached maximum iterations without terminating... either there's a problem/bug or you can fix by raising the max iterations or lowering Tol_deC")
    return intersection_list

此外,以防有人需要,我编写了 Casteljau 算法的代码,用于拆分任意次数的贝塞尔曲线。在上述代码中,我只是用 halveCubic 替换了它,这只是将 Casteljau 方法显式化并限制为三次情况(t=0.5)。
def splitBezier(points,t):
#returns 2 tuples of control points for the two resulting Bezier curves
    points_left=[]
    points_right=[]
    (points_left,points_right) =  splitBezier_deCasteljau_recursion((points_left,points_right),points,t)
    points_right.reverse()
    return (points_left,points_right)
def splitBezier_deCasteljau_recursion(cub_lr,points,t):
    (cub_left,cub_right)=cub_lr
    if len(points)==1:
        cub_left.append(points[0])
        cub_right.append(points[0])
    else:
        n = len(points)-1
        newPoints=[None]*n
        cub_left.append(points[0])
        cub_right.append(points[n])
        for i in range(n):
            newPoints[i] = (1-t)*points[i] + t*points[i+1]
        (cub_left, cub_right) = splitBezier_deCasteljau_recursion((cub_left,cub_right),newPoints,t)
    return (cub_left, cub_right)

我希望你能帮助某些人,感谢您的帮助,Yves!

0

简化系统怎么样?

通过适当的线性组合,您可以消除s^3t^3中的一个,并解决剩余的二次方程。通过将结果插入另一个方程式中,您可以得到一个单一未知数的单一方程式。

或者解决通过结果获得的代数方程:

36011610661302281 - 177140805507270756*s - 201454039857766711*s^2 + 1540826307929388607*s^3 + 257712262726095899*s^4 - 4599101672917940010*s^5 + 1114665205197856508*s^6 + 6093758014794453276*s^7 - 5443785088068396888*s^8 + 1347614193395309112*s^9 = 0

(http://www.dr-mikes-maths.com/polynomial-reduction.html)


我相信我对这种方法不太熟悉,我会去了解一下。那么你认为这个方法可以用于找到任意两个三次(特别是贝塞尔)曲线的交点吗? - mathandy
是的,它确实可以,但我没注意到这是贝塞尔曲线的交集。对于这个问题,有更合适的方法,可以利用凸壳性质来解决。 - user1196549

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