Return

Question

Hello,

I struggled with this, can someone help me understand how to view this return*?

def is_even(x):
if x == 0:
return True
else:
return is_odd(x-1)

def is_odd(x):
return not is_even(x)

print(is_odd(17))
print(is_even(23))

Answer 1

Let’s start with return not is even (x) please

Answer 2

Hi!

It recursively negates the result of the result of the result...of (x - 1) == 0, until x = 0.

Example with 3 (odd):
3 == 0 [True] [False] -> False -> not (x - 1) ->
2 == 0 [True] [False] -> not False -> (not) not (x - 1) ->
1 == 0 [True] [False] -> not not False -> (not, not) not (x - 1) ->
0 == 0 [True] [False] -> not not not True -> False

Actually, there is no need of is_odd(x) function.

def is_even(x):
if x == 0:
return True
return not is_even(x - 1)

print(is_even(17)) # False (odd)
print(is_even(18)) # True (even)

Answer 3

Thanks guys,

I am still struggling with this, is there an easier example please?

Answer 4

I think that it will be easier for you to understand it, if you remove the is_odd function, like in my example.
It works the same way, without it.

not is_even means that if x = 0, then you return True in the is_even function. But since there is 'not' in front of it, it becomes False.

Assume that you call is_even(17), then 17 == 0 is False, so it returns is_odd(17 - 1), which returns not is_even, which checks 16 == 0 (False), so again it returns is_odd(x - 1), which returns not is_even(15). And here is the tricky part. Since you are going deeper in the recursion tree, while x > 0, the not operator is stacked (you got always not not not not not) and when x = 0 you return True, but it might be False in the end:
x = 3
3 == 0 -> returns is_odd(x - 1) -> returns not is_even(2) ->
2 == 0 -> returns is_odd(x - 1) ->
returns not is_even(1) ->
1 == 0 -> returns is_odd(x - 1) ->
returns not is_even(0) ->
0 == 0 -> returns True.

Now count how many times you call not is_even(x). 3 times, right?
So, not not not True = False (3 is odd)
1. not True = False
2. not False = True
3. not True = False

Hope it's more clear now.

Answer 5

Thanks a lot Boris, I believe I get it now. Thanks for patiently explaining!!!

Answer 6

I'm glad if helped.
Perhaps this approach is just for recursion demo. It works for 0 <= x <= 997.