python - 尝试通过二分搜索查找非完美立方体的立方根时出现无限循环

Question

这段代码在 deltanum = 0.0000000000001 时给出了相当准确的结果，但在 deltanum = 0.00000000000001 时进入无限循环(向 deltanum 添加另一个零).

它只发生在非完美的立方体上，它适用于像 1000 这样的完美立方体。为什么？

我是编程新手，跟随 OSSU。

num = 100
high = num
low = 0
icount = 0
cuberoot = (high + low)/2      #cuberoot of num
deltanum = 0.00000000000001
while abs(cuberoot**3 - num)>=deltanum:
    icount+=1
    print(icount)
    if cuberoot**3 > num:
        high = cuberoot
    elif cuberoot**3 < num:
        low = cuberoot
    else:
        break
    cuberoot = (high + low)/2
print("Cube root: " + str(cuberoot))
print("Number of iterations: " + str(icount))

Answer 1

您正在使用 float。 float 数学在精度方面存在缺陷 - 您的 delta 可能太小而无法正常工作，并且您的解决方案在值之间翻转而从未达到您的 while 条件限制。参见 Is floating point math broken?有关为什么 float 有时会“损坏”的更多原因。

您也可以将其限制为一定的重复次数:

num = 100
high = num
low = 0
icount = 0
maxcount = 100000
cuberoot = (high + low)/2      #cuberoot of num
deltanum = 0.00000000000001
while abs(cuberoot**3 - num)>=deltanum:
    icount+=1
    print(icount)
    if cuberoot**3 > num:
        high = cuberoot
    elif cuberoot**3 < num:
        low = cuberoot
    else:
        break
    cuberoot = (high + low)/2

    if icount > maxcount:
        print("Unstable solution reached after ",maxcount, "tries")
        break
print("Cube root: " + str(cuberoot) + " yields " + str(cuberoot**3))
print("Number of iterations: " + str(icount))

输出:

Unstable solution reached after  100000 tries
Cube root: 4.641588833612779 yields 100.00000000000003
Number of iterations: 100001