Recursion in Python return None

Question

Why does the function integerCubeRootHelper return None instead of an integer value?

def integerCubeRootHelper(n, left, right):
    cube = lambda x: x * x * x # anonymous function to cube a number
    assert(n >= 1)
    assert(left < right)
    assert(left >= 0)
    assert(right < n)
    assert(cube(left) < n), f'{left}, {right}'
    assert(cube(right) > n), f'{left}, {right}'
    
    mid=(left+right)//2
    if ( left == mid ):
        return left
    elif(cube(mid)> n):
        integerCubeRootHelper(n, left, mid)
    else:
        integerCubeRootHelper(n, mid, right)

def integerCubeRoot(n):
    if (n == 1): 
        return 1
    if (n == 2):
        return 1
    return integerCubeRootHelper(n, 0, n-1)

integerCubeRoot(3)

Questions must be in English. You can post in Portuguese at [pt.so]. — John Kugelman, Jan 26 '22 at 00:33
When you make a recursive call to `integerCubeRootHelper()`, you are throwing away its result rather than `return`ing it. — jasonharper, Jan 26 '22 at 01:13

score 0 · Answer 1 · answered Jan 26 '22 at 01:13

You don't return the result of the recursive function - you only run it.

If you change your code to the following you will no longer get None:

def integerCubeRootHelper(n, left, right):
    cube = lambda x: x * x * x # anonymous function to cube a number
    assert(n >= 1)
    assert(left < right)
    assert(left >= 0)
    assert(right < n)
    assert(cube(left) < n), f'{left}, {right}'
    assert(cube(right) > n), f'{left}, {right}'
    
    mid=(left+right)//2
    if ( left == mid ):
        return left
    elif(cube(mid)> n):
        return integerCubeRootHelper(n, left, mid) # CHANGED
    else:
        return integerCubeRootHelper(n, mid, right) # CHANGED

def integerCubeRoot(n):
    if (n == 1): 
        return 1
    if (n == 2):
        return 1
    return integerCubeRootHelper(n, 0, n-1)

ans = integerCubeRoot(3)
print(ans) # 1

Oleksii Tambovtsev · Answer 2 · 2022-01-26T01:19:54.543

The main problem is you miss return:

def integerCubeRootHelper(n, left, right):
    cube = lambda x: x * x * x  # anonymous function to cube a number
    assert (n >= 1)
    assert (left < right)
    assert (left >= 0)
    assert (right < n)
    assert (cube(left) < n), f'{left}, {right}'
    assert (cube(right) > n), f'{left}, {right}'

    mid = (left + right) // 2
    if (left == mid):
        return left
    elif (cube(mid) > n):
        return integerCubeRootHelper(n, left, mid)
    else:
        return integerCubeRootHelper(n, mid, right)


def integerCubeRoot(n):
    if (n == 1):
        return 1
    if (n == 2):
        return 1
    return integerCubeRootHelper(n, 0, n - 1)


print(integerCubeRoot(3))

And some more advice about your code that can help you to increase code correctness and readability:

Your functions' name should be lowercase. Link
You have to remove redundant parentheses.
Always use a def statement instead of an assignment statement that binds a lambda expression directly to a name. Link

Full PEP 8 style guide

Updated version:

def integer_cube_root_helper(n, left, right):
    def cube(x):
        return x * x * x
    
    assert n >= 1
    assert left < right
    assert left >= 0
    assert right < n
    assert cube(left) < n, f'{left}, {right}'
    assert cube(right) > n, f'{left}, {right}'

    mid = (left + right) // 2
    if left == mid:
        return left
    elif cube(mid) > n:
        return integer_cube_root_helper(n, left, mid)
    else:
        return integer_cube_root_helper(n, mid, right)


def integer_cube_root(n):
    if n == 1:
        return 1
    if n == 2:
        return 1
    return integer_cube_root_helper(n, 0, n - 1)


print(integer_cube_root(3))

Recursion in Python return None

2 Answers2