Question

我目前正在研究python的polyfit库。我确实有一个特定的输入值A和一个期望的输出值B。我想生成一个复杂度为5 <= n <= 10的随机多项式，当给定A作为输入时，将B作为解决方案。最好的方法是什么？

Answer 1

正如我在评论中建议的那样，如果只需要拟合一个值，则可以为所有系数选择任意值，然后调整最终系数，以使曲线穿过所需的点。从根本上讲，这等效于在图形上绘制曲线，然后向上或向下滑动直到与所需的点相交。

import random

def generate_polynomial(degree, a, b):
    """chooses coefficients for a polynomial of the given degree, such that f(a) == b"""

    #to fit only one data point, we can choose arbitrary values for every coefficient except one, which we initially set to zero.
    coefficients = [0] + [random.randint(1, 10) for _ in range(degree-1)]

    #now calculate f(a). This will probably not be equal to b, initially.
    y = sum(coefficient * a**n for n, coefficient in enumerate(coefficients))

    #setting the final coefficient to their difference will cause f(a) to equal b.
    coefficients[0] = b - y

    return coefficients

seq = generate_polynomial(3, 4, 42)
print(seq)

一个可能的结果：

[-18, 7, 2]

这对应于f(x) == 2*x^2 + 7*x - 18。手动确认f（4）== 2 * 16 + 7 * 4-18 = 42很容易。

为随机输入生成随机多项式

1 个答案: