sicp-exercises/ex-2-20.rkt

#lang sicp
;; 2.17 : while I was having fun with this,
;; I came up with the following monstrosity
;; yes, it works, and is completely equivalent
;; the second, less horrible implementation
(define (last-pair l)
  (or (and (or (null? l) (null? (cdr l))) l)
      (last-pair (cdr l))))
(define (last-pair2 l)
  (cond
    ((null? l) l)
    ((null? (cdr l)) l)
    (#t (last-pair2 (cdr l)))))

;; 2.18
(define (reverse l)
  (define (iter acc l)
    (if (null? l)
        acc
        (iter (cons (car l) acc)
              (cdr l))))
  (iter '() l))

(equal? '(4 3 2 1) (reverse '(1 2 3 4)))

;; 2.19

(define (cc amount coin-values)
  (define (first-denomination cv)
    (car cv))
  (define (except-first-denomination cv)
    (cdr cv))
  (define no-more? null?)
  (cond ((= amount 0) 1)
        ((or (< amount 0) (no-more? coin-values)) 0)
        (else
         (+ (cc amount
                (except-first-denomination
                 coin-values))
            (cc (- amount
                   (first-denomination
                    coin-values))
                coin-values)))))

;; 2.20
(define (filter f l)
  (cond
    ((null? l) l)
    ((f (car l)) (cons (car l) (filter f (cdr l))))
    (#t (filter f (cdr l)))))
(define (same-parity a . b)
  (cons a
        (filter (if (even? a) even? odd?)
                b)))
(equal? (same-parity 1 2 3 4 5 6) '(1 3 5))

;; 2.21
(define (square x) (* x x))
(define (square-list-direct items)
  (if (null? items)
      '()
      (cons (square (car items)) (square-list-direct (cdr items)))))
(define (square-list-map items)
  (map square items))

;; 2.22 this exercise asks for an explanation on why the
;; accumulated result is in reverse order.
;; This is because, at each iteration, the newest item
;; is added to the head of the list - always.
;; this means that the first processed item will be added
;; to the head, then the second, then the third. So
;; the list '(1 2 3) will be processed as such:

;; '(1 2 3) '()
;; '(2 3) '(1)
;; '(3) '(4 1)
;; '() '(9 4 1)

;; So the lists behave like stacks, i.e. FILO ADTs.

;; the "solution" proposed by Louis Reasoner
;; does not work, because that doesn't change the order in
;; which the items are processed or where the items are
;; inserted. It only changes the positions of car and cdr,
;; and in practice this will ruin the structure of the list,
;; nothing more.

;; 2.23
;; a trivial implementation: (just use map and discard result)
(define (trivial-for-each f l)
  (map f l)
  #t)
;; but this isn't much useful, as the memory is still allocated
;; (if promptly recollected by the GC) for map, defeating the purpose.
;; Here's an implementation that simply ignores the results:
(define (my-for-each f l)
  (if (null? l)
      #t
      (begin (f (car l))
             (my-for-each f (cdr l)))))
;; no consing is done, and this generates an iterative process,
;; so no stack allocations either. Equivalent to a for-loop in
;; a language like C or C++.


;; 2.25 - can also be written as: (car (cdaddr l))
(car (cdr (car (cdr (cdr '(1 3 (5 7) 9))))))
(car (car '((7))))
; third one's too long. I'll be using cadr to mean (car (cdr x))
(cadr (cadr (cadr (cadr (cadr (cadr '(1 (2 (3 (4 (5 (6 7))))))))))))

;; 2.26 - doesn't require doing anything, just see what they do lol


;; 2.27
;; similar to regular reverse, with the addition that
;; if the current element is itself a list, we cons
;; its reverse instead.
;; O(n) time complexity (where n = leaf count)
;; O(k) space complexity (where k = branch depth? not sure about the correct terminology)
(define (deep-reverse l)
  (define (iter acc l)
    (cond
      ((null? l) acc)
      ((list? (car l)) (iter (cons (deep-reverse (car l)) acc)
                             (cdr l)))
      (#t (iter (cons (car l) acc)
               (cdr l)))))
  (iter '() l))

;; 2.28
;; ooh, this one's spicy
;; okay, I think I'll first use a recursive process for this,
;; then translate it to an iterative one.
;; Note: iterative process will require a reverse at the end,
;; unless mutable conses are in play. That's why I used sicp lang for this.
(define (fringe l)
  (cond
    ((null? l) l)
    ((list? (car l)) (append (fringe (car l))
                             (fringe (cdr l))))
    (#t (cons (car l)
              (fringe (cdr l))))))

(define (fringe-iter l)
  "This is mostly the same actually. We just change the mechanism for adding stuff.
   Efficiency-wise, the only difference is that this impl only goes another level
   deep for every level of nested lists (instead of going another level deep *at every iteration*)
   So, better, but still not perfect if you have a tree a billion levels deep.

   The difference here from a typical iterative process is the mutation: instead
   of using pure functions I just had the iterator keep track of the end of the list."
  (define (add end x)
    (let ((c (cons x '())))
      (set-cdr! end c)
      c))
  (define (iter acc-end l)
    (cond
      ((null? l) acc-end)
      ((list? (car l)) (iter (iter acc-end (car l))
                             (cdr l)))
      (#t (iter (add acc-end (car l))
                (cdr l)))))
  (let ((sentinel (cons #f '())))
    (iter sentinel l)
    (cdr sentinel)))