python编程错误RecursionError:比较中超出了最大递归深度

Question

这是我的python程序，也是练习6.4的答案。

#!/usr/bin/env python3

"""
Exercise 6.4.
A number, a, is a power of b if it is divisible by b and a/b is a power of b.
Write a function called is_power that takes parameters a and b and returns
True if a is a power of b.
Note: you will have to think about the base case.
"""

def is_power(a, b):
    """Checks if a is power of b."""
    if a == b:
        return True
    elif a%b == 0:
        return is_power(a/b, b)
    else:
        return False

print("is_power(10, 2) returns: ", is_power(10, 2))
print("is_power(27, 3) returns: ", is_power(27, 3))
print("is_power(1, 1)  returns: ",  is_power(1, 1))
print("is_power(10, 1) returns: ", is_power(10, 1))
print("is_power(3, 3)  returns: ",  is_power(3, 3))

每当我一次又一次地尝试显示这个错误时，我都会得到这个错误。请给我指点一下我的程序哪里出了错。

is_power(10, 2) returns:  False
is_power(27, 3) returns:  True
is_power(1, 1)  returns:  True
Traceback (most recent call last):
  File "C:\Users\Aaban Shakeel\Desktop\Think-Python-2e---my-solutions-master\ex6\ex6.4.py", line 23, in <module>
    print("is_power(10, 1) returns: ", is_power(10, 1))
  File "C:\Users\Aaban Shakeel\Desktop\Think-Python-2e---my-solutions-master\ex6\ex6.4.py", line 16, in is_power
    return is_power(a/b, b)
  File "C:\Users\Aaban Shakeel\Desktop\Think-Python-2e---my-solutions-master\ex6\ex6.4.py", line 16, in is_power
    return is_power(a/b, b)
  File "C:\Users\Aaban Shakeel\Desktop\Think-Python-2e---my-solutions-master\ex6\ex6.4.py", line 16, in is_power
    return is_power(a/b, b)
  [Previous line repeated 1021 more times]
  File "C:\Users\Aaban Shakeel\Desktop\Think-Python-2e---my-solutions-master\ex6\ex6.4.py", line 13, in is_power
    if a == b:
RecursionError: maximum recursion depth exceeded in comparison

Answer 1

似乎迭代地重写算法比使用递归更好。这是因为尾递归在python中效率不高。更多信息：https://stackoverflow.com/a/3323013/5746085

Answer 2

这是因为对于b=1，您的递归将永远不会结束-我会补充：

def is_power(a, b):
    """Checks if a is power of b."""
    if a == b:
        return True
    elif b==1:
        return False
    elif a%b == 0:
        return is_power(a/b, b)
    else:
        return False

Answer 3

您必须为数字1添加额外的检查，因为您的算法在此行中停滞(参数'a‘停止更改)

is_power(a / b, b)

如果b ==为1，则永远调用相同的函数(实际上直到达到最大递归深度:P )

is_power(10 / 1, 1)
is_power(10 / 1, 1)
...

递归限制可以防止堆栈溢出。