def find_ratio()

in scripts/util.py [0:0]

• 30 lines of code
• 11 McCabe index (conditional complexity)

def find_ratio(a, n, s, r0=None):
    """given the start (a), count (n), and sum (s), find the ratio required"""
    if n == 2:
        return s / a - 1
    an = a * n
    if n == 1 or s == an:
        return 1
    if r0 is None:
        # we only need to know which side of 1 to guess, and the iteration will work
        r0 = 1.1 if an < s else 0.9

    # geometric sum formula
    def f(r):
        return a * (1 - r**n) / (1 - r) - s

    # derivative of f
    def df(r):
        rm1 = r - 1
        rn = r**n
        return (a * (rn * (n * rm1 - r) + r)) / (r * rm1**2)

    MIN_DR = 0.0001  # negligible change
    r = r0
    # print(f"r(0)={r0}")
    MAX_TRIES = 64
    for i in range(1, MAX_TRIES + 1):
        try:
            dr = f(r) / df(r)
        except ZeroDivisionError:
            log.error(f"Failed to find ratio due to zero division! Returning r={r0}")
            return r0
        r = r - dr
        # print(f"r({i})={r}")
        # if the change in r is small, we are close enough
        if abs(dr) < MIN_DR:
            break
    else:
        log.error(f"Could not find ratio after {MAX_TRIES}! Returning r={r0}")
        return r0
    return r