Just one minor thing, I don´t understand why procedure right-branch is like this:
(define (right-branch mobile)
(car (cdr mobile)))
instead of:
(define (right-branch mobile)
(cdr mobile))
As far as I know, car procedure return the first item in a list and cdr returns the "rest" of the list. So, if we have a mobile like this:
((1 2) (3 4))
the left branch is (1 2) and the right branch is (3 4). Is like this or am I missing something? Thanks.
And the same for procedure branch-structure.
The issue here is that (cdr '((1 2) (3 4))) is not '(3 4). Rather, it is '((3 4)).
This is because the list is a list, containing three elements: two lists and the empty list (null). That means the list, expressed as pairs, actually looks like this:
'((1 2) . ((3 4) . ()))
Since cdr gets the right side of a pair, it returns the rest of the list, so even if only one element is left, it is wrapped in a containing list.