Scheme: SICPの練習問題(2.17～2.28)

最近SICPを読み始めた(オンラインなので、英語。きつい・・・)
SICPはMITが出してる本で、バイブル的な何からしい。
1章の問題は数学チックなのが多いので、半分くらい飛ばしてしまった。後でやろう。
折角なので、(自分の)答えを載せてみたり。

下のは職場のGaucheで作って(おいおい)、家のMzSchemeで確認。
MzSchemeだと「()」が使えないので、「'()」に変更。
一つMzSchemeで動かないのがあった。処理系依存なコードだったのかしら？

; 2.17
(define (last-pair list)
  (define (last-pair-itr val list2)
    (if (null? list2)
 val
      (last-pair-itr (car list2) (cdr list2))))
  (last-pair-itr (car list)(cdr list)))
(last-pair '(1 2 3 4 5 6))

; 2.18
(define (reverse list)
  (define (reverse-itr ret list2)
    (if (null? list2)
 ret
      (reverse-itr (cons (car list2) ret) (cdr list2))))
 (reverse-itr '() list))
(reverse '(1 2 3 4 5 6))


; 2.21
(define (square-list-no-map list)
  (if (null? list)
      '()
    (cons (* (car list) (car list))
   (square-list-no-map (cdr list)))))
(square-list-no-map '(1 2 3 4 5 6))



(define (square-list list)
  (map (lambda (x)(* x x)) list))
(square-list '(1 2 3 4))


; 2.22
(define (square-list items)
  (define (square x)(* x x))
  (define (iter things answer)
    (if (null? things)
 answer
      (cons (square (car things))
     (iter (cdr things) answer))))
  (iter items '()))
(square-list '(1 2 3 4 5))


; 2.23
(define (for-each proc items)
  (if (null? items)
      #f 
    (and (proc (car items))
  (for-each proc (cdr items)))))

; 2.25
(car (cdr (car (cdr (cdr '(1 3 (5 7) 9))))))
(car (car '((7))))
(car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr '(1 (2 (3 (4 (5 (6 7))))))))))))))))))

; 2.26
(define x (list 1 2 3))
(define y (list 4 5 6))
(append x y)
(cons x y)
(list x y)

; 2.27
(define x (list (list 1 (list 2 3) (list 4 5)) (list 6 7 8)))
(reverse x)
(define (deep-reverse list)
  (define (reverse-itr ret list)
    (cond 
     ((null? list) ret)
     ((pair? (car list))
      (reverse-itr (cons (reverse-itr '() (car list)) ret) (cdr list)))
     (else (reverse-itr (cons (car list) ret)
   (cdr list)))))
  (reverse-itr '() list))

; Gaucheでは動いたコード
;(define (deep-reverse list)
;  (cond
;   ((null? list) '())
;   ((pair? (car list)) (cons (car (cdr (car list)))(car (car list))))
;   (else (cons (deep-reverse (cdr list)) (car list)))))
(deep-reverse x)


; 2.28
(define (fringe l2)
  (cond ((null? l2) '())
 ((not (pair? l2)) (cons l2 '()))
 (else (append (fringe (car l2))
        (fringe (cdr l2))))))
(fringe (list x x))