How to Get My Recursive Function to Return 0 in Python

Question

For my recursion assignment, the only built in function we are allowed to use is len() and 'print()'. So, not even int() is allowed to be used. The function I am supposed to make is supposed to break apart a number and print out each digit in the number. So,

>>breakApart(757890)

7
5
7
8
9
0

This is what I have come up with so far,

def breakApart(number):
    c=0
    while number//10**c>=10:
        c+=1
    integer_part=((number-(number%10**c))//(10**c))
    if number//10**c==0:
        return
    elif number//10**c>0:
        print(integer_part)
        return breakApart(number%10**c)

This works for all numbers, except it will not print out 0s. I was thinking of writing something along the lines of,

if (number%10**c)>+10 and (number%10**c)//(10**(c-1))<1:
    print(0)
    return breakApart(number%10**c)

But if there are consecutive 0s, it will not print all of them. I guess I could use another counter like I did at the beginning, but if anyone could think of less convoluted ways for me to approach this, please let me know.

@juanpa.arrivillaga "For my recursion assignment..." - sounds like it. — Tom Wyllie, Jul 26 '17 at 16:41
@sneakysnake I'm assuming using `print` is also allowed, given that you've used it? — Tom Wyllie, Jul 26 '17 at 16:42
What's the purpose of having len() if you can't use the str() function? Do you have to pass in an int? This would be really simple with recursion and a string — Easton Bornemeier, Jul 26 '17 at 16:45
You can use len if the input is a string, but you can't change the data type of int(4) to str(4) for example — , Jul 26 '17 at 16:47
And no, it cannot print 0s because the remainder from the mod ignores all of those 0s @coldspeed — , Jul 26 '17 at 16:51
@EastonBornemeier it's pretty simple with `int` and recursion as well. — juanpa.arrivillaga, Jul 26 '17 at 16:55
@juanpa.arrivillaga I agree, but can't argue the fact that simply pulling characters off the sides of the string isn't a simple method if it was allowed — Easton Bornemeier, Jul 26 '17 at 16:58

score 3 · Answer 1 · answered Jul 26 '17 at 16:47

3

Use recursion to your advantage here, making the print call come after your recursive call:

In [1]: def tell_digits(n):
   ...:     rest, first = n // 10, n % 10
   ...:     if rest:
   ...:         tell_digits(rest)
   ...:     print(first)
   ...:

In [2]: tell_digits(757890)
7
5
7
8
9
0

Note what happens if I put the print call before the recursive call:

In [3]: def tell_digits(n):
   ...:     rest, first = n // 10, n % 10
   ...:     print(first)
   ...:     if rest:
   ...:         tell_digits(rest)
   ...:

In [4]: tell_digits(757890)
0
9
8
7
5
7

answered Jul 26 '17 at 16:47

juanpa.arrivillaga

88,713
10
131
172

Thank you very much! I don't want to steal this answer, however, since it deviates quite a bit from my code, but I'll use it as an example – Jul 26 '17 at 16:53
Also, I'm a bit of a beginner, so I am trying to understand your code. What does "if rest" mean? If rest is what? – Jul 26 '17 at 17:02
1

@sneakysnake `if rest:` checks if `rest` is [*falsey*](https://stackoverflow.com/questions/39983695/what-is-truthy-and-falsy-in-python-how-is-it-different-from-true-and-false). Also, see the [docs](https://docs.python.org/3/library/stdtypes.html#truth-value-testing). In this case, it is the same as `if rest != 0:` – juanpa.arrivillaga Jul 26 '17 at 17:05

Easton Bornemeier · Answer 2 · 2017-07-26T16:50:28.363

0

If number can be a string:

def breakApart(number):
    if len(number) == 0:
        return
    else:
        print(number[0])
        breakApart(number[1:])

Or reverse it:

def breakApart(number):
    if len(number) == 0:
        return
    else:
        breakApart(number[:-1])
        print (number[-1])

edited Jul 26 '17 at 16:50

answered Jul 26 '17 at 16:48

Easton Bornemeier

1,918
8
22

No, we were given an example of the input being an integer type, not string type. – Jul 26 '17 at 16:50

score 0 · Answer 3 · answered Jul 27 '17 at 18:28

0

Something to keep you up at night:

>>> def breakApart(number):
...     return number > 9 and breakApart(number // 10) or print(number % 10)
... 
>>> breakApart(757890)
7
5
7
8
9
0
>>>

Python 3 specific, of course.

answered Jul 27 '17 at 18:28

cdlane

40,441
5
32
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This might take me a little while to understand, but thank you for another interesting answer :) – Jul 28 '17 at 13:36

How to Get My Recursive Function to Return 0 in Python

3 Answers3