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)