Yun的算法

Question

我想尝试实现Yun的多项式无平方因子分解算法.来自维基百科(f是多项式):

a0 = gcd(f, f'); b1 = f/a0; c1 = f'/a0; d1 = c1 - b1'; i = 1
repeat
ai = gcd(bi, di); bi+1 = bi/ai; ci+1 = di/ai; i = i + 1; di = ci - bi'
until b = 1

但是,我不确定第二步.我想将它用于具有整数系数的多项式(不需要monic或primitive).是否有可能b1 = f/a0只使用整数实现除法？

我找到了合成分部的代码:

def extended_synthetic_division(dividend, divisor):
    '''Fast polynomial division by using Extended Synthetic Division. Also works with non-monic polynomials.'''
    # dividend and divisor are both polynomials, which are here simply lists of coefficients. Eg: x^2 + 3x + 5 will be represented as [1, 3, 5]

    out = list(dividend) # Copy the dividend
    normalizer = divisor[0]
    for i in xrange(len(dividend)-(len(divisor)-1)):
        out[i] /= normalizer # for general polynomial division (when polynomials are non-monic),
                             # we need to normalize by dividing the coefficient with the divisor's first coefficient
        coef = out[i]
        if coef != 0: # useless to multiply if coef is 0
            for j in xrange(1, len(divisor)): # in synthetic division, we always skip the first coefficient of the divisor,
                                              # because it is only used to normalize the dividend coefficients
                out[i + j] += -divisor[j] * coef

    # The resulting out contains both the quotient and the remainder, the remainder being the size of the divisor (the remainder
    # has necessarily the same degree as the divisor since it is what we couldn't divide from the dividend), so we compute the index
    # where this separation is, and return the quotient and remainder.
    separator = -(len(divisor)-1)
    return out[:separator], out[separator:] # return quotient, remainder.

对我来说问题是out[i] /= normalizer.它会一直适用于Yun的整数(floor)分区b1 = f/a0吗？它是否总是可以划分f/gcd(f, f')？是out[separator:](余)总是会为零？

Answer 1

事实上，“除法p/GCD(p, p')始终有效（即“精确”，Z 中没有余数） ”是从 的定义得出的GCD。对于任何多项式p及其q整除GCD(p,q)都是p且q精确的。这就是为什么它被称为GCD最大公约数：

和的最大公约数是一个多项式，它能整除和，使得和的每个公约数也能整除。pqdpqpqd

PS 在更专业的https://math.stackexchange.com/上提出这样的纯数学问题更有意义