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racket（plt scheme升级版） 下载地址：http://racket-lang.org/download/

 

习题：1.3 定义一个过程，它以三个数作为参数，返回其中两个数之和

 

#lang racket
(define (min x y)
  (if (> x y)
      y
      x)
  )

(define (mul-min x y z)
  (if (> (min x y) z)
      z
      (min x y)))

(define (add-two-bigger x y z)
  (- (+ x y z)
     (mul-min x y z)))

(add-two-bigger -3 -6 3)
      

 

习题：1.5 正则序和应用序 解释情况

答：应用序 p一直递归，死循环。 正则序可以输出0

 

 

例子1：牛顿法求平方根

 

#lang racket  
 
(define (square x)
  (* x x))

(define (improve guess x)
 (/ (+ (/ x guess) guess) 2)
  )

(define (enough?  guess x)
  (> 0.0001
       (abs(- (square guess) x))))

(define (sqrt-iter guess x)
  (if (enough? guess x)
      guess
      (sqrt-iter (improve guess x) 
                 x))
       )
(define (sqrt x)
  (sqrt-iter 1.0 x))

(sqrt 4)
      

 

习题1.6 创造new-if来代替if方法。

 

#lang racket  
 
(define (square x)
  (* x x))

(define (improve guess x)
 (/ (+ (/ x guess) guess) 2)
  )

(define (enough?  guess x)
  (> 0.0001
       (abs(- (square guess) x))))
(define (new-if p then-clause else-clause)
  (cond (p then-clause)
        (else else-clause)))

(define (sqrt-iter guess x)
  (new-if (enough? guess x)
      guess
      (sqrt-iter (improve guess x) 
                 x))
       )
(define (sqrt x)
  (sqrt-iter 1.0 x))

(sqrt 4)
      

  答：出现死循环，原因是：解释器使用应用序进行解析代码，对于求值描述是首先对运算符和各个运算对象求值，而后将得到的过程应用于得到的实际参数。所以对于new-if中的方法，他会先去计算p then-clause else-clause全部计算好，然后再执行cond方法，这个导致了else-clause出现了递归，一直重复计算。

 

习题1.7 更改enough方法

 

#lang racket
(define (sqrt3 x)
  (sqrt3-iter 1.0 x 0.1))

(define (sqrt3-iter guess x last-guess)
  (if (enough? guess last-guess)
      guess
      (sqrt3-iter (improve guess x) x guess)))

(define (enough? guess last-guess)
  (> 0.00001 (/ (abs (- guess last-guess)) guess)))

(define (improve guess x)
  (/ (+ (/ x (* guess guess)) (* 2 guess)) 3))

(sqrt3 2000000000000000000000007)

 

把这个与1.8代码做对比，确实时间上差很多

 

习题1.8 求立方根

 

 

#lang racket
(define (sqrt3 x)
  (sqrt3-iter 1.0 x))

(define (sqrt3-iter guess x)
  (if (enough? guess x)
      guess
      (sqrt3-iter (improve guess x) x)))

(define (enough? guess x)
  (> 0.00001 (abs (- (* guess guess guess) x))))

(define (improve guess x)
  (/ (+ (/ x (* guess guess)) (* 2 guess)) 3))

(sqrt3 27)