how the return is working in recursion with reference to the following code

Question

def subs(l):
    if l == []:
        return [[]]

    x = subs(l[1:])

    return x + [[l[0]] + y for y in x]

1) I am trying to understand working of the above code

2) If l = [1,2,3] at the end of the recursion we will get an empty list of list

3) In last next iteration we will be given the result to the x will be 3 (I am thinking it will go to the next step which is return)

4) In these returns I am thinking it will add return x + [[l[0]] + y for y in x] should return (3,3).

5) By ending of these step the answer should be (3,3).

The final answer according to be at the second step must be [[],[3,3]].

But the code is printing a different answer.

o/p 

[], [3]]

Can any one explain where my assumption went wrong and how return x + [[l[0]] + y for y in x] these steps are working?

Can't reproduce. Your code works perfectly for me... (assuming it's supposed to return the list of sublists). — Julien
– Julien, Commented Mar 12, 2019 at 3:45
I think you made an error in computing the value of x. When l=[3], then x=[[]]. In step 4, the result of return x + [[l[0]] + y for y in x] will be [[]] + [[3] + []] which is [[], [3]] — Nitin Labhishetty
– Nitin Labhishetty, Commented Mar 12, 2019 at 3:49
code is working fine but i am unable to understand how ? according my assumption the o/p should be different i am unable to figure out the return working in the above code — mdvenkatesh nuhk
– mdvenkatesh nuhk, Commented Mar 12, 2019 at 4:52

user10325516 · Accepted Answer · 2019-03-12 07:26:12Z

I made some modifications to the code to make it print intermediate results:

def subs(l, num=0):

    print(' ' * num, 'current l is:', l)
    print()

    if l == []:
        return [[]]

    x = subs(l[1:], num + 4)
    print(' ' * num, 'current x is:', x)

    answer = x + [[l[0]] + y for y in x]
    print(' ' * num, 'current answer is:', answer)

    print()

    return answer

subs([1, 2, 3])

Let's have a look at the output (it's an image):

Indentation depictures resursion depth. Black arrows show calculation flow. Red arrows show which l and x correspond to each other (belong to the same namespace). From the image one can see how recursion differs from loops. Recursion goes all the way down to the trivial case and then goes up. Pay attention to the red arrows.

So, first, x is never 3. It just can't be actually. Second, l can be [3]. Corresponding x is [[]] returned by the final recursion step (trivial case). These values give you answer [[], [3]]. And this is x for the previous l = [2, 3].

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