云算法

Question

我想尝试用云算法实现多项式的无平方分解.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

然而，我不确定第二步。我想用它作为整数系数的多项式(不是必要的单元或本原)。用整数实现除法b1 = f/a0是可能的吗？

我找到了合成师的代码

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。它是否总是与云的b1 = f/a0的整数(地板)部门一起工作？是否总是可以对f/gcd(f, f')进行分割？out[separator:] (余数)总是为零吗？

Answer 1

“p/GCD(p, p')中的除法将始终工作(即”精确“，在Z中没有余数)这一事实遵循于GCD的定义。对于任何多项式p和q，它们的GCD(p,q)都精确地将p和q分开。这就是为什么它被称为GCD，即最伟大

最大公共除数 of p和q是一个多项式d，它将p和q除以，使得p和q的每个公共除数也除以d。

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