SICP Exercise 2.22: Reversed Squares

Louis Reasoner tries to rewrite the first square-list procedure of exercise 2.21 so that it evolves an iterative process:

(define (square-list items)
  (define (iter things answer)
    (if (null? things)
        answer
        (iter (cdr things) 
              (cons (square (car things))
                    answer))))
  (iter items nil))

Unfortunately, defining square-list this way produces the answer list in the reverse order of the one desired. Why?

Louis then tries to fix his bug by interchanging the arguments to cons:

(define (square-list items)
  (define (iter things answer)
    (if (null? things)
        answer
        (iter (cdr things)
              (cons answer
                    (square (car things))))))
  (iter items nil))

This doesn't work either. Explain.

Back in exercise 2.18 we defined the following procedure for reversing a list:

(define (reverse l)
  (define (iter remaining result)
    (if (null? remaining)
        result
        (iter (cdr remaining) (cons (car remaining) result))))
  (iter l nil))

Now looking carefully at Louis' first iterative implementation of square-list we can see that it is practically identical to our implementation of reverse. Ignoring the different names and formatting, it differs only in the first argument passed to cons when building up the result and equivalent answer lists. In the case of reverse it simply uses (car remaining). However, for Louis' first iterative implementation of square-list it uses (square (car things)). I.e. it squares the car of the remaining elements to process, rather than just using the car of the remaining list directly.

Based on this we can say that what Louis has done is to simply extend the reverse procedure so that it squares each value during the reversal procedure. He's starting with an empty list as the result and then going through the list to square starting with the head, squaring the current value and putting the result onto the head of the result list. So the square of the first element becomes the last item in the list, the square of the second element becomes the second-last item in the list and so on.

Louis' second attempt at producing an iterative implementation of square-list exchanges the two arguments to cons. While this corrects the ordering of the squares it no longer produces a list as the result.

What we find instead is that answer, the result list build up so far, is used as the first argument, while (square (car things)), the square of the current element, is used as the second argument. Now cons only builds a pair - it does not guarantee to produce a validly constructed list. To produce a valid list, although not necessarily a flattened list, from cons the second argument must be itself a valid list. Instead what Louis' procedure does is to return a pair, the second element of which is the square of the last item in the list, while the first element in the pair is itself a pair, with the second element being the square of the second-last item in the list and the first element being a pair... And this nesting of pairs in the first element continues until we reach a pair in which the first element is the empty list and the second element is the square of the first element in the list.

Of course a picture's worth a thousand words, so let's see it in action:

> (square-list (list 1 2 3 4 5 6 7 8 9 10))
((((((((((() . 1) . 4) . 9) . 16) . 25) . 36) . 49) . 64) . 81) . 100)

If you look at this structure closely what you'll see that the pairs that are constructed are a sort of "reverse list structure". A list is constructed of pairs in which the second element of each pair in the list is itself a pair, except for the last pair where it is nil. The values held in the list are the first elements of each pair. What Louis' implementation has constructed is a data structure composed of pairs in which the first element of each pair is itself a pair, except for the first pair where it is nil. In Louis' structure the values are the second elements of each pair.