SICP_3.15-3.20

3.15

对于z1

z1 ------>[ * ][ * ]

                |     |

               v     v

              [ * ][ * ]----->[ * ][ ]

                |                  |

               v                  v

               wow             b

对于z2

z2 ----->[ * ][ * ]

               |     |                            

               |     + ------------------>[ * ][ * ]------>[ * ][ ]

               |                                  |                    |

               |                                  v                   v

               |                                  a                   b

               |                                                       ^

               |                                                       |

               +----------------------->[ * ][ * ]------>[ * ][ ]

                                                  |

                                                  v

                                                  wow

3.16

(define l2 (list 1))
(define l3 (cons l2 l2))        ->结果为4
(define l4 (list l3))


(define l1 '(1 2 3))  ->结果为3

(define l5 (list a))
(define l6 (cons l5 l5))
(define l7 (cons l6 l6))         ->结果为7

 (define l8 (set-car! (cddr l1) l1)) ->什么都不返回

3.17

(define new-count-pairs
  (let ((store '()))
    (lambda (x)
      (cond ((not (pair? x)) 0)
            ((memq x store) 0)
            (else (set! store (cons x store))
                  (+ (new-count-pairs (car x))
                     (new-count-pairs (cdr x))
                     1))))))

把检查过的元素放到一个表中。在计数之前检查是否存在。注意memq的用法！！！

3.18

这一题的一些解法

(define (cycle? x)
  (define (iter search-list seen)
    (cond ((not (pair? search-list)) #f)
          ((memq search-list seen) #t)
          (else (or (iter (car search-list) (cons search-list seen))
                    (iter (cdr search-list) (cons search-list seen))))))
  (iter x '()))

这一题重点是要注意car也可以形成环

另一种貌似有问题的解法

他这里的解法没有考虑表中出现重复元素的情况。

3.19

(define (loop? ls)
  (define (iter x y)
    (let ((x-walk (list-walks 1 x))
          (y-walk (list-walks 2 y)))
      (cond ((or (null? x-walks) (null? y-walks)) #f)
            ((eq? x-walks y-walks) #t)
            (else (iter x-walk y-walk)))))
  (iter ls ls))

(define (list-walks step ls)
  (cond ((null? ls) '())
        ((zero? step) ls)
        (else (list-walks (- step 1)
                          (cdr ls)))))

不会这题，参考，用了算法：龟兔赛跑

3.20

还是理解的不透彻