1.46 迭代式改进的一般 形式.

iterative-improve.scm

; 迭代式改进. -- 哈哈哈. 最终的社区精简版, 竟然是我这种写法. --- 编程理念的差异. 在可读性和技巧性的取舍上, 我优先选择 可读性.
(define (iterative-improve good-enough? impove)
    (define (f x) 
        (if (good-enough? x)
            x
            (f (impove  x))
        )
    )
    f
)

; 社区改进版: 直接递归调用函数自己.
(define (iterative-improve good-enough? improve) 
   (lambda (guess) 
     (if (good-enough? guess) 
         guess 
         ((iterative-improve good-enough? improve) (improve guess))))) 

; old sqrt from 1.7
(define (sqrt x)
    (sqrt-iter 1.0 x)
)

(define (sqrt-iter guess x)
    (if (good-enough? guess x)
        guess
        (sqrt-iter (impove guess x) x)
    )
)

(define (impove guess x)
    (average guess (/ x guess))
)

(define (average x y)
    (/ (+ x y) 2)
)

(define (good-enough? guess x)
    (< (abs (- (square guess) x)) 0.01)
)

(define (square x) (* x x))

; rewrite sqrt-iter.
(define (sqrt-iter guess x)
    ((iterative-improve (lambda (y) (good-enough? y x)) 
                      (lambda (y) (impove y x)))
      guess
    )
)

; rewrite fixed-point. base 1.35.scm
(define (fixed-point f first-guess)
    (define pre-guess first-guess)
    ((iterative-improve (lambda (guess) 
                         (define old-pre-guess pre-guess)
                         (set! pre-guess guess)
                         (< (abs (- old-pre-guess guess)) tolerance))
                       f)
     (f first-guess)
    )
)

(define (golden-ratio)
    (fixed-point (lambda (x) (+ 1 (/ 1 x))) 1.0)
)

#|
> (golden-ratio)
1.6180327868852458
|#

; 需要记录 前一个值...  社区有更好的方法吗?  ==》 社区 “偷懒” 的方法. 但是这样计算结果的,会是 guess, 而不是 next. 和原实现的结果不同.
(define (fixed-point f first-guess) 
   ((iterative-improve 
     (lambda (x) (close-enough? x (f x))) 
     f) 
    first-guess))

#|
> (golden-ratio)
1.6180371352785146
|#

社区答案, 表现一般.