使用递归确定半完美数

Question

I am in need of your help again. I'm writing a code that determines whether a number is a semiperfect number or not by returning a boolean value. So first I thought I would make a list of the factors of the number excluding the number itself

def isSemiPerfect(n):
factor = []
for i in range(n-1, 0, -1):
    if n%i == 0:
        factor.append(i)

If I were to check for the number 12, it would return

factor = [1, 2, 3, 4, 6]

Then I need to make a code to use recursion to check if you add certain numbers it would be equal to 12

True = 6 + 4 + 2 or 6 + 3 + 2 + 1

Can someone tell me how I can use recursion to do trial and error? Like I'll always start with the biggest number and try a path with the next biggest number until I've tried all combinations.

I know there isn't a lot to go on with I just hope that you can tell me how I can use recursion effectively. Thank you!

Answer 1

You can think about it this way.

The question "Can [1,2,3,4,6] sum to 12"?

Is the same as "Can [2,3,4,6] sum to 11" or "Can [2,3,4,6] sum to 12"?

One uses the first element (and has lower sum) and the other does not use the first element and has the same sum.

so a start function would be:

def f(lst,sum_left):
    return f(lst[1:], sum_left-lst[0]) or f(lst[1:],sum_left)

However, this function does not know when to stop. So we need some base cases .

Obviously, if sum is 0, the answer is trivially yes:

if sum_left == 0:
    return true

Another base case is, if the sum is < 0, we have taken a too big element and we can never recover.

if sum_left < 0:
    return true

Also, we can risk running out of elements like [1,2] can never sum to 50, so we add a base case if there are no elements in the list:

if not lst:
    return false

Putting it all together:

def f(lst, left):
    if left == 0:
        return True
    if left < 0:
        return False
    if not lst:
        return False
    
    return f(lst[1:],left-lst[0]) or f(lst[1:],left)