在DrRacket(或Scheme)中使用此格式形成树

时间:2017-09-27 19:21:03

标签: functional-programming scheme binary-tree racket

所以我使用DrRacket并将结构定义为:

(define-struct thing (a b))

然后,我有一个这种格式的示例树:

(define v (make-thing 
              (make-thing (make-thing 1 2) 
                          (make-thing 3 4))  
              (make-thing (make-thing 5 6) 
                          (make-thing 7 8))))

我需要构建一个函数,它接收一个非正数,也是2的幂(如1,2,4,8,16 ......),并输出数字1(如果n = 1)或制作一个'采用上述格式。

到目前为止,我已经尝试了很多方法,但我最终得到的结果如下:

(make-thing (make-thing (make-thing 1 1) 
                        (make-thing 1 1)) 
            (make-thing (make-thing 1 1) 
                        (make-thing 1 1))))

我无法弄清楚如何正确增加变量 - 我已经能够弄清楚如何应用所需数量的make-thing

这是我现在使用的代码:

(define (helper pow current)
  (cond
    ((= pow 0) 1)
    (else (make-thing 
             (helper (- pow 1) current) 
             (helper (- pow 1) (- current 1))))))

(define current 1)

(define (build-thing-or-number n)
  (cond
    ((= n 1) 1)
    (else (make-thing 
             (helper (- (log n 2) 1) current) 
             (helper (- (log n 2) 1) (add1 current))))))

2 个答案:

答案 0 :(得分:3)

请允许我致电build-thing-or-number g,这样我就可以打字了。

所以你想要

(g 1)   =   1                  ; you've got this covered already
(g 2)   =   (thing 1 2)
(g 4)   =   (thing (thing 1 2) (thing 3 4))
(g 8)   =   (thing (thing (thing 1 2) (thing 3 4))       ; 1..4
                   (thing (thing 5 6) (thing 7 8)))      ; 5..8

我们看到,如果我们有更多参数,那么更容易做到:

(define (h a b)                        ; from ... to
  (cond 
     ((= (+ a 1) b)                    ; like (h 1 2):
        (make-thing a b))              ;    just make the thing
     (else                             ; e.g. (h 1 4)   (h 5 8)
        (let ([c (/ (+ a b -1) 2)])    ;       c = 2     c = 6 
          (make-thing                  ; c is the middle
                 (h a c)               ; make the first half   1..2   1..4
                 (h ..... ))))))       ; and the second        3..4   5..8

我们将其简单地用作

(define (g b)
  (cond ((= b 1) 1)
        (else (h ..... ))))

那就是那个!

答案 1 :(得分:2)

#lang racket

(define-struct thing (a b) #:transparent)

我已将事物定义为透明,以便于验证

(define (helper level max)
  (if (equal? (/ level 2) 1)
      (make-thing (- max 1) max)
      (make-thing (helper (/ level 2) (- max (/ level 2))) (helper (/ level 2) max))))

max等于当前事物中的最高数字 level是同一形状中最简单(读取最低值)事物中的最高数字(即(事物(事物5 6)(事物7 8))最大值为8,等级为4)

(define (build-thing-or-number n)
  (cond
    ((= n 1) 1)
    (else (helper n n))))

正如您所看到的,帮助者完成所有工作。

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