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(define (make-list-ft from to)
(if (> from to)
'()
(cons from (make-list-ft (1+ from) to))))
;; Exercise 1.3 (proper lists)
;;? Program own version of list?, thus proper-list?
;; This is done by running to the end of list, always checking if pair.
;; If last element is '() its a proper list, otherwize its not.
(define (proper-list? list)
(if (pair? list)
(proper-list? (cdr list))
(null? list)))
;;? Can you write your predicate without using if or cond?
;; Hmm i cant see how this would be done, as we are using recursion which requires a stop condition.
;;? Is your function efficient? What is the time complexity?
;; This runs over all the elements in the list, so O(n).
;; Hmm reading through the solution, i see that we can use other conditional functions like `or` and `and`.
;; Lets try that
(define (proper-list? list)
(or (null? list) (and (pair? list) (proper-list? (cdr list)))))
;; Okay thats just his solution
;; Exercise 1.5 (every second element)
;; Hmm the text mentions that we could write it with (every-nth-element)
(define (every-nth-element n list)
(let recur ((list list) (index 0))
(cond ((null? list) '())
((zero? (modulo index n)) (cons (car list) (recur (cdr list) (1+ index))))
(else (recur (cdr list) (1+ index))))))
(define (every-2th-element list)
(every-nth-element 2 list))
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