如何修复我的代码？ 使用递归，返回一个包含所有 k 的列表，使得 3 ≤ k ≤ n 并且 k 可以被 3 或 5 整除，但不能同时被 5 整除？

Question

So I am reviewing for my class, and as the title suggests I am trying to return a list from a recursive function that includes the positive values divisible by 3 or 5 but not both. So for example, div_by_3_xor_5(20) should return [3, 5, 9, 10, 12, 18, 20] , with 15 being missing because it is divisible by both 3 and 5.

My approach to figuring out recursive solutions is usually to write it as a regular loop first, then see what I need to tweak to turn it into a recursive function. The regular loop function of this is simple enough:

def div_by_3_xor_5(n):
    a_list = []
    for i in range(3, n + 1):
        div_by_3 = (i % 3 == 0)
        div_by_5 = (i % 5 == 0)
        
        if div_by_3 and div_by_5:
            pass
        elif div_by_3 or div_by_5:
            a_list.append(i)
    return a_list

# correct output for div_by_3_xor_5(20): [3, 5, 9, 10, 12, 18, 20]

I have tried converting it into a recursive function but the output is wrong.

def div_by_3_xor_5(n, k = 3):
    if k >= n:
        return [n]
    
    for i in range(k, n + 1):
        div_by_3 = (i % 3 == 0)
        div_by_5 = (i % 5 == 0)
        
        if div_by_3 and div_by_5:
            pass
        elif div_by_3 or div_by_5:
            return [i] + div_by_3_xor_5(n, k + 1)
    
# incorrect output for div_by_3_xor_5(20): 
# [3, 5, 5, 6, 9, 9, 9, 10, 12, 12, 18, 18, 18, 18, 18, 18, 20, 20]

Could anyone help me figure out what's going wrong, so that I return [3, 5, 6, 9, 10, 12, 18, 20] instead?

Answer 1

def div_by_3_xor_5(n, k = 3):
    if k >= n:
        return [n]
    
    div_by_3 = (k % 3 == 0)
    div_by_5 = (k % 5 == 0)
        
    if div_by_3 and div_by_5:
        return div_by_3_xor_5(n, k + 1)
    elif div_by_3 or div_by_5:
        return [k] + div_by_3_xor_5(n, k + 1)
    else:
        return div_by_3_xor_5(n, k + 1)

# output of div_by_3_xor_5(20) was correct: [3, 5, 6, 9, 10, 12, 18, 20]

I figured it out after looking at my code again thanks to Karl's comment. I was trying to loop over the integers not divisible by either 3 or 5, but I could've done that with a recursive call to the next integer. Thank you Karl!

Answer 2

You somehow need to make the recursion act on a smaller piece of the problem each time.

Here's a divide-and-conquer form of recursion, although it's a real stretch to force-fit recursion into this requirement in my view.

def wacky_numbers(hi, lo=3):
    if hi == lo:
        test3 = lo%3 == 0
        test5 = lo%5 == 0
        if test3 != test5:
            return [lo]
        else:
            return []
    else:
        mid = (hi+1+lo)//2
        return wacky_numbers(mid-1,lo) + wacky_numbers(hi,mid)

For extra protection you might want to test hi<lo