练习 1.44

根据题目给出的定义，写出平滑函数 smooth ：

;;; 44-smooth.scm

(define dx 0.00001)

(define (smooth f)
    (lambda (x)
        (/ (+ (f (- x dx))
              (f x)
              (f (+ x dx)))
            3)))

测试：

1 ]=> ((smooth square) 5)

;Value: 25.000000000066663

smooth-n-times

题目还要求我们写出一个可以连续进行 n 次平滑的函数（叫它 smooth-n-times 好了）。

根据描述，表达式 (smooth-n-times f 3) 的展开式应该为：

(smooth-n-times f 3)

(smooth (smooth-n-times f 2))

(smooth (smooth (smooth-n-times f 1)))

(smooth (smooth (smooth f)))

根据这个展开式，最简单直观的 smooth-n-times 可以这样定义：

;;; 44-smooth-n-times.scm

(load "44-smooth.scm")

(define (smooth-n-times f n)
    (if (= n 0)
        f
        (smooth (smooth-n-times f (- n 1)))))

上面的 smooth-n-times 是一个递归计算过程， smooth-n-times 的迭代计算过程定义如下：

;;; 44-iter-smooth-n-times.scm

(load "44-smooth.scm")

(define (smooth-n-times f n)
    (define (iter i smoothed-f)
        (if (= i 0)
            smoothed-f
            (iter (- i 1)
                  (smooth smoothed-f))))
    (iter n f))

使用 repeated 定义 smooth-n-times

事实上，在前面的 smooth-n-times 定义里，我们做了重复的工作，因为在 练习 1.43 里，我们就写过生成连续调用序列的函数 repeated ，而 smooth-n-times 里对 smooth 的连续调用完全可以使用 repeated 来完成，从而避免显式的递归，以及 off-by-one 错误：

(repeated smooth 3)

(lambda (f)
    (smooth (repeated smooth 2)))

(lambda (f)
    (smooth (smooth (repeated smooth 1))))

(lambda (f)
    (smooth (smooth (smooth f))))

以下是使用 repeated 实现的 smooth-n-times 的定义：

;;; 44-smooth-n-times-using-repeated.scm

(load "44-smooth.scm")
(load "43-repeated.scm")

(define (smooth-n-times f n)
    (let ((n-times-smooth (repeated smooth n)))
        (n-times-smooth f)))

测试

无 repeated ，递归计算的 smooth-n-times ：

1 ]=> (load "44-smooth-n-times.scm")

;Loading "44-smooth-n-times.scm"...
;  Loading "44-smooth.scm"... done
;... done
;Value: smooth-n-times

1 ]=> ((smooth-n-times square 10) 5)

;Value: 25.000000000666663

无 repeated ，迭代计算的 smooth-n-times ：

1 ]=> (load "44-iter-smooth-n-times.scm")

;Loading "44-iter-smooth-n-times.scm"...
;  Loading "44-smooth.scm"... done
;... done
;Value: smooth-n-times

1 ]=> ((smooth-n-times square 10) 5)

;Value: 25.000000000666663

带 repeated 的 smooth-n-times （是递归计算还是迭代计算取决于所使用的 repeated 函数）：

1 ]=> (load "44-smooth-n-times-using-repeated.scm")

;Loading "44-smooth-n-times-using-repeated.scm"...
;  Loading "44-smooth.scm"... done
;  Loading "43-repeated.scm"... done
;... done
;Value: smooth-n-times

1 ]=> ((smooth-n-times square 10) 5)

;Value: 25.000000000666663

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