递归函数查找同时被三个数字整除的最小数字

Question

I am trying to find the smallest integer that is divisible by 2, 3, and 5 through a recursive function as follows:

def recursiva(n):

    lista = []
    if(n%2==0 and n%3==0 and n%10==0):
        lista.append(n)

    n = n - 1
    recursiva(n)

    return min(lista)

recursiva(100)

But even for small numbers like 100 I am having stack overflow, as seen in the error message:

RecursionError: maximum recursion depth exceeded in comparison

I wonder:

• What am I doing wrong?
• Is there any way, instead of passing a fixed value per parameter, to make the function look for the smallest number divisible by 2,3 and 5 within the set of integers?

Answer 1

To fix it, before calling recursiva() again, check if n is equal or less than 0 ( n<=0 ). Without a condition, it will not stop, and will loop infinitely.

Answer 2

In a recursive function you need to have at least 2 possible return statements, 1 where some end condition is met and the answer is returned and 1 where what is returned is new call to the function with some modified input.

Additionally for your problem you need a method of saving the results of the previous calls somewhere. You might consider passing the list of successful calls as a parameter to the function.

Answer 3

I got it like this:

def recursiva(n = 1):
    if (n%2==0 and n%3==0 and n%10==0):
        return n
    else:
        return recursiva(n+1)
recursiva()