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+ 2

Can anyone explain this recursion code!!

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)) output: True False

22nd Aug 2018, 6:58 PM
Mominul Islam Rasel
Mominul Islam Rasel - avatar
3 Answers
+ 10
The `is_odd` function is simply returning the inverse (`not` keyword) of the `is_even` function, and that is logically correct. The recursion is happening in the `is_even` function, so let's break it down in parts: (1) The base case of the recursion is when x = 0. When that happens, `is_even` returns True and the recursion ends. (2) From (1) we conclude that x cannot be negative or you'll end up in an infinite recursive "loop". That's because we are passing x-1 to the `is_odd` function when the base case isn't met, which means the arguments passed to the functions are decreasing. If x is initially negative and it only decreases, then x = 0 never happens and the recursion doesn't end. (3) If x > 0 the `is_even` function takes the logical assumption that for any x that is even, x-1 is obligatory odd. That is not hard to prove mathematically. So by all means, what happens when you call, say, `is_odd(5)`, is that initially x=5 and it keeps getting passed from one function to another, until it x=0.
22nd Aug 2018, 8:11 PM
Eduardo Petry
Eduardo Petry - avatar
+ 5
take a pencil and a sheet of paper and try with 5 and 4. 4 is even? check if 3 is not even check if 2 is even check if 1 is not even check if 0 is even returns true so 1 is not even so 2 is even so 3 is not even so 4 is even.
22nd Aug 2018, 8:13 PM
Oma Falk
Oma Falk - avatar
+ 1
is_even returns the same logical value as it receives. is_odd returns the opposite, it reverses the outcome. is_even is the only one that decrements x (x = x - 1). The base case x=0 terminates returning True. Look at the trivial cases first is_even(0) returns True is_odd(0) returns the opposite =False (one reverse) is_even(1) returns the same as is_odd(0) =False (one reverse) is_odd(1) returns the opposite = True (two reverses cancel out) is_even(2) returns the same as is_odd(1) = True (2 reverses) is_even(x) If x is any positive* number it will go through an even number of reverses (=True) if it's even and an odd number (=False) if it's odd. (*Yes this program fails to check for negative values and will cause a runtime error for those) is_odd(x) Reverses the outcome of is_even(x) and so produces False for even and True for odd.
5th Nov 2018, 9:18 AM
Ian Berry
Ian Berry - avatar