Question

这种Durand-Kerner算法（here）的实现有什么问题？

def durand_kerner(poly, start=complex(.4, .9), epsilon=10**-16):#float('-inf')):
    roots = []
    for e in xrange(poly.degree):
        roots.append(start ** e)
    while True:
        new = []
        for i, r in enumerate(roots):
            new_r = r - (poly(r))/(reduce(operator.mul, [(r - r_1) for j, r_1 in enumerate(roots) if i != j]))
            new.append(new_r)
        if all(n == roots[i] or abs(n - roots[i]) < epsilon for i, n in enumerate(new)):
            return new
        roots = new

当我尝试时，我必须用KeyboardInterrupt来阻止它，因为它不会停止！ poly是pypol库的多项式实例。

提前谢谢你，魔方

编辑：使用numpy多项式需要9次迭代：

In [1]: import numpy as np

In [2]: roots.d1(np.poly1d([1, -3, 3, -5]))
3
[(1.3607734793516519+2.0222302921553128j), (-1.3982133295376746-0.69356635962504309j), (3.0374398501860234-1.3286639325302696j)]
[(0.98096328371966801+1.3474626910848715j), (-0.3352519326012724-0.64406860772816388j), (2.3542886488816044-0.70339408335670761j)]
[(0.31718054925650596+0.93649454851955749j), (0.49001572078718736-0.9661410790307261j), (2.1928037299563066+0.029646530511168612j)]
[(0.20901563897345796+1.5727420147652911j), (0.041206038662691125-1.5275192097633465j), (2.7497783223638508-0.045222805001944255j)]
[(0.21297050700971876+1.3948274731404162j), (0.18467846583682396-1.3845653821841168j), (2.6023510271534573-0.010262090956299326j)]
[(0.20653075193800668+1.374878742771485j), (0.20600107336130213-1.3746529207714699j), (2.5874681747006911-0.00022582200001499547j)]
[(0.20629950692533283+1.3747296033941407j), (0.20629947661265013-1.374729584400741j), (2.5874010164620169-1.899339978055233e-08j)]
[(0.20629947401589896+1.3747296369986031j), (0.20629947401590082-1.3747296369986042j), (2.5874010519682002+9.1830687539942581e-16j)]
[(0.20629947401590029+1.3747296369986026j), (0.20629947401590026-1.3747296369986026j), (2.5874010519681994+1.1832913578315177e-30j)]
Out[2]: 
[(0.20629947401590029+1.3747296369986026j),
 (0.20629947401590029-1.3747296369986026j),
 (2.5874010519681994+0j)]

使用pypol多项式它永远不会完成（它可能是pypol中的一个错误）：

In [3]: roots.d2(poly1d([1, -3, 3, -5]))
^C---------------------------------------------------------------------------
KeyboardInterrupt

但我找不到错误!!

EDIT2 ：将__call__方法与Martin's Poly进行比较：

>>> p = Poly(-5, 3, -3, 1)
>>> from pypol import poly1d
>>> p2 = poly1d([1, -3, 3, -5])

>>> for i in xrange(-100000, 100000):
    assert p(i) == p2(i)


>>>
>>> for i in xrange(-10000, 10000):
    assert p(complex(1, i)) == p2(complex(1, i))


>>> for i in xrange(-10000, 10000):
    assert p(complex(i, i)) == p2(complex(i, i))


>>>

EDIT3 ：如果根不是复数，pypol工作正常：

In [1]: p = pypol.funcs.from_roots([4, -2, 443, -11212])

In [2]: durand_kerner(p)
Out[2]: [(4+0j), (443+0j), (-2+0j), (-11212+0j)]

所以只有当根是复数时它才有效！

EDIT4 ：我为numpy多项式编写了一个稍微不同的实现，并发现在一次迭代后，（维基百科多项式的）根是不同的：

In [4]: d1(numpyp.poly1d([1, -3, 3, -5]))
Out[4]: 
[(0.98096328371966801+1.3474626910848715j),
 (-0.3352519326012724-0.64406860772816388j),
 (2.3542886488816044-0.70339408335670761j)]

In [5]: d2(pypol.poly1d([1, -3, 3, -5]))
Out[5]: 
[(0.9809632837196679+1.3474626910848717j),
 (-0.33525193260127306-0.64406860772816377j),
 (2.3542886488816048-0.70339408335670772j)] ## here

EDIT5 ：嘿！如果我将行if all(n == roots[i] ... )更改为if all(str(n) == str(roots[i]) ... )，它就会完成并返回正确的根!!!“

In [9]: p = pypol.poly1d([1, -3, 3, -5])

In [10]: roots.durand_kerner(p)
Out[10]: 
[(0.20629947401590029+1.3747296369986026j),
 (0.20629947401590013-1.3747296369986026j),
 (2.5874010519681994+0j)]

但问题是：为什么它适用于不同的复数比较？

更新
现在它有效，我做了一些测试：

In [1]: p = pypol.poly1d([1, -3, 3, -1])

In [2]: p
Out[2]: + x^3 - 3x^2 + 3x - 1

In [3]: pypol.roots.cubic(p)
Out[3]: (1.0, 1.0, 1.0)

In [4]: durand_kerner(p)
Out[4]: 
((1+0j),
 (1.0000002484566535-2.708692281244913e-17j),
 (0.99999975147728026+2.9792265510301965e-17j))

In [5]: q = x ** 3 - 1

In [6]: q
Out[6]: + x^3 - 1

In [7]: pypol.roots.cubic(q)
Out[7]: (1.0, (-0.5+0.8660254037844386j), (-0.5-0.8660254037844386j))

In [8]: durand_kerner(q)
Out[8]: ((1+0j), (-0.5-0.8660254037844386j), (-0.5+0.8660254037844386j))

Answer 1

您的算法看起来很好，对于我来说，它适用于维基百科中的示例

import operator
class Poly:
    def __init__(self, *koeff):
        self.koeff = koeff
        self.degree = len(koeff)-1

    def __call__(self, val):
        res = 0
        x = 1
        for k in self.koeff:
            res += x*k
            x *= val
        return res

def durand_kerner(poly, start=complex(.4, .9), epsilon=10**-16):#float('-inf')):
    roots = []
    for e in xrange(poly.degree):
        roots.append(start ** e)
    while True:
        new = []
        for i, r in enumerate(roots):
            new_r = r - (poly(r))/(reduce(operator.mul, [(r - r_1) 
                                     for j, r_1 in enumerate(roots) if i != j]))
            new.append(new_r)
        if all((n == roots[i] or abs(n - roots[i]) < epsilon) for i, n in enumerate(new)):
            return new
        roots = new

print durand_kerner(Poly(-5,3,-3,1))

给出

[(0.20629947401590026+1.3747296369986026j), 
 (0.20629947401590026-1.3747296369986026j), 
 (2.5874010519681994+8.6361685550944446e-78j)]

Answer 2

关于你的“编辑5”：这是因为str（）没有将数字格式化为完全精确度。

>>> print str((2.5874010519681994+8.636168555094445e-78j))
(2.58740105197+8.63616855509e-78j)
>>> print repr((2.5874010519681994+8.636168555094445e-78j))
(2.5874010519681994+8.636168555094445e-78j)
>>>

所以不要这样做。

在任何情况下，代码中的相等测试：

if all(n == roots[i] or abs(n - roots[i]) < epsilon for i, n in enumerate(new)):

是多余的;如果n == roots[i]，那么abs(n - roots[i])将为零，所以你可以只做

if all(abs(n - roots[i]) < epsilon for i, n in enumerate(new)):

并付出一些努力来确定epsilon的默认值;正如我在评论中指出的那样，解决X**3 == 1收敛但你的默认epsilon太小而无法阻止它永远循环。对于默认的epsilon，1.12e-16看起来更好。

为了娱乐，尝试不合适的Poly（[ - 1,3，-3,1]）......三个根等于（1 + 0j）...它需要超过600次迭代，并且错误在最后10次左右的迭代几乎惊人地跳了起来，直到它从左边的远处到达一个合理的解决方案。

Durand-kerner实施不起作用

2 个答案: