Why does my recursive function return None?

问题

I'm currently trying to wrap my head around learning Python and I've come to a bit of a stall on recursive functions. In Think Python, one of the exercises is to write a function that determines if number a is a power of number b using the following definition:

"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."

The current state of my function is:

def isPower(a,b):
    return a % b == 0 and (a/b) % b == 0

print isPower(num1,num2)

As it is, this produces the result I expect. However the chapter is focused on writing recursive functions to reduce redundancy and I'm not quite sure how I can turn the final "(a/b) % b == 0" into a recursion. I've attempted:

def isPower(a,b):
    if a % b != 0:
        return False
    elif isPower((a/b),b):
        return True

But that just returns None.

What is the proper way to recurse this function?

回答1:

You are forgetting a base case, when a == 1:

def isPower(a,b):
    if a == 1:
        return True
    if a % b != 0:
        return False
    elif isPower((a/b),b):
        return True
    else
        return False

However this has some other problems - if a is 0 then it will never finish and if b is 0 then you will get a divide-by-zero.

Here is a verbose solution that as far as I can tell will work for all integer combinations:

def isPower(a,b):
    if a == 0 or b == 0:
        return False
    def realIsPower(a, b):
        if a == 1:
            return True
        elif a%b != 0:
            return False
        elif realIsPower((a/b), b):
            return True
        else:
            return False
    return realIsPower(a, b)

EDIT: My code didn't work for cases when both a and b are negative. I'm now comparing their absolute values.

EDIT2: Silly me, x^0 == 1, so a == 1 should ALWAYS return true. That also means I don't have to compare a to b before the recursion. Thanks @Javier.

回答2:

you need an additional case, for when both conditionals return false

def isPower(a,b):
    if a % b != 0:
        return False
    elif isPower((a/b),b):
        return True
    else
        return False

回答3:

def isPower (a,b):
    return a==1 or (a!=0 and b!=0 and b!=1 and isPower((a/b),b))

回答4:

Here is my code. From what I tested, it works:

def ispower(a, b):

    if b == 1 or b == 0:
        return False
    if b <= 0 or a <= 0:
        return False
    if a % b == 0:
        if ((a / b) / b) == 1:
            return True
        else:
            return ispower(a / b, b)
    else:
        return False
        print ispower(-10, 2)

回答5:

try this,

def ispower(a,b):
  if b==a:
    return True
  elif a<b:
    return False
  else:
    return ispower(a*1.0/b, b)

回答6:

def is_power (a, b):

    if a == 1:
        return True
    if a == 0 and b == 0:
        return True
    if a == 0 or b == 0:
        return False
    if a%b != 0:
        return False
    elif is_power ((a/b), b):
        return True

回答7:

Here's my answer, it's a little bit cleaner:

def is_power(a, b):
    if a == 1:
        return True
    if a == 0 or b == 0:
        return False
    if a % b == 0 and is_power(a/b, b):
        return True
    else:
        return False

来源：https://stackoverflow.com/questions/3719853/why-does-my-recursive-function-return-none

标签

python

recursion