I think the trick here is to imagine how you can transform your input into the code that you'd want to use to compute the sum. Let's write each of your inputs in the fully expanded form, in terms of cons
and '()
and whatever other atoms appear in your data:
'() == '()
'(()) == (cons '() '())
'(1 1) == (cons 1 (cons 1 '()))
'(1 (1 1) 1) == (cons 1 (cons 1 (cons 1 '())) (cons 1 '()))
'(1 (1 (1 1)) 1) == ...
Now, look what would happen if you replaced each occurrence of cons
with +
, and each occurrence of '()
with 0
, and each occurrence of something that's not '()
with 1
. You'd have:
'() => 0 == 0
(cons '() '()) => (+ 0 0) == 0
(cons 1 (cons 1 '())) => (+ 1 (+ 1 0)) == 2
(cons 1 (cons 1 (cons 1 '())) (cons 1 '())) => (+ 1 (+ 1 (+ 1 0)) (+ 1 0)) == 4
... => ... == ...
Notice that those sums are exactly the values that you want! Based on this, it seems like you might not want to treat your input as a list so much as a tree built from cons cells. In general, you can map over a tree by specifying a function to apply to the recursive results of processing a pair, and a function to process the atoms of the tree:
(define (treeduce pair-fn atom-fn tree)
(if (pair? tree)
(pair-fn (treeduce pair-fn atom-fn (car tree))
(treeduce pair-fn atom-fn (cdr tree)))
(atom-fn tree)))
You could then implement that mapping of cons
to +
and everything else to 1
if it's a list and 0
if it's not by:
(define (non-null-atoms tree)
(treeduce +
(lambda (atom)
(if (not (null? atom))
1
0))
tree))
This yields the kinds of results you'd expect:
(non-null-atoms '()) ;=> 0
(non-null-atoms '(())) ;=> 0
(non-null-atoms '(1 1)) ;=> 2
(non-null-atoms '(1 (1 1) 1)) ;=> 4
(non-null-atoms '(1 (1 (1 1)) 1)) ;=> 5
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