poly_purple - 5 months ago 37

Python Question

I'm currently studying python from http://www.sololearn.com/Play/Python and I'm having trouble understanding why this code works.

`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(1))

I get how recursion works for a fibonacci and factorial but I can't wrap my head around this one.

Answer

`is_even'`

s base case resolves to `True`

. Since `is_odd(x)`

returns `not is_even(x)`

, the value `True`

will be a part of the expression returned by `is_odd`

. The question is how many times will that `True`

value be *negated*. By tracing the calls you can see it will be negated an even number of times, and hence "retain" its truthiness, when `x`

is odd [e.g.: `x=3`

==> `(not (not (not (not True))))`

== `True`

] and an odd number of times, and hence "lose" its truthiness, when `x`

is even [e.g.: `x=2`

==> `(not (not (not True)))`

== `False`

]. There's probably some term from logic that names this general property of multiple negation.