(load "shared/util.scm")

(display "ex-2.17\n")

(define (last-pair a)
  (if (null? (cdr a))
    (car a)
    (last-pair (cdr a))))

(define a (list 1 2 3 4))
(assert (last-pair a) 4)

(display "\nex-2.18\n")

; Recursive solution probably less efficient because of calls to append and
; list.
(define (reverse l)
  (if (null? l)
    l
    (append (reverse (cdr l)) (list (car l)))))

; Iterative solution. Probably much more efficient.
(define (reverse l)
  (define (list-iter acc l)
    (if (null? l)
      acc
      (list-iter (cons (car l) acc) (cdr l))))
  (list-iter (list) l))

(display (reverse a))

(display "\n\nex-2.19\n")

(define us-coins (list 50 25 10 5 1))
(define uk-coins (list 100 50 20 10 5 2 1 0.5))

(define no-more? null?)
(define first-denomination car)
(define except-first-denomination cdr)

(define (cc amount coin-values)
  (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)))))

(display (cc 100 us-coins)) (newline)

; The order of the values in the list do not change the result because the
; procedure tries all potential combinations of using and not using the coins.
(display (cc 100 (reverse us-coins)))

(display "\n\nex-2.20\n")

;  (define (same-parity? a b) (= (modulo a 2) (modulo b 2)))

(define (same-parity x . xs)
  (define (same-parity-iter parity-first rest)
    (cond ((null? rest) (list))
          ((= parity-first (modulo (car rest) 2)) (cons (car rest) (same-parity-iter parity-first (cdr rest))))
          (else (same-parity-iter parity-first (cdr rest)))))
  (same-parity-iter (modulo x 2) (cons x xs)))

(display (same-parity -1 2 3 4 5 6 7)) (newline)
(display (same-parity 2 3 -4 5 6 7)) (newline)

(display "\nex-2.21\n")

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

(display (square-list a)) (newline)

(define (square-list items)
  (map (lambda (x) (* x x)) items))

(display (square-list a)) (newline)

(display "\nex-2.22\n")

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

; Louis adds the first element (squared) to the accumulator for each iteration
; step. That means later elements are prepended later, hence the order of the
; list is reversed.

(display (square-list a)) (newline)

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

(display (square-list a)) (newline)

; Changing the arguments does not work either, because it results in building a
; reverse list of lists, e.g.: ((((nil 16) 9) 4) 1)
; The only solution would be to use append which is inefficient.

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

(display (square-list a)) (newline)

(display "\nex-2.23")

(define (for-each proc xs)
  (cond
    ((null? xs) ())
    (else (proc (car xs)) (for-each proc (cdr xs)))))

(for-each (lambda (x) (newline) (display x))
          (list 57 321 88))
(newline)