sicp习题试解 (1.31)

王朝other·作者佚名 2006-01-09

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; ======================================================================

;

; Structure and Interpretation of Computer Programs

; (trial answer to excercises)

;

; 计算机程序的构造和解释(习题试解)

;

; created: code17 03/05/05

; modified:

; (保持内容完整不变前提下，可以任意转载)

; ======================================================================

;; SICP No.1.31

;; 递归版本

(define (product-r term a next b)

(if (> a b)

(* (term a)

(product-r term (next a) next b))))

;; 迭代版本

(define (product-i term a next b)

(define (iter a result)

(if (> a b)

result

(iter (next a) (* result (term a)))))

(iter a 1))

;; 阶乘

(define (factorial n)

(define (term i) i)

(define (next i) (+ i 1))

(product-i term 1 next n))

;; 求pi

(define (john-wallis-pi n)

(define (term i) (/ (* (* 2 i) (* 2 (+ i 1)))

(square (+ 1 (* 2 i)))))

(define (next i) (+ i 1))

(* (product-i term 1 next n) 4.0))

;; Test-it:

;; > (factorial 5)

;; 120

;; > (factorial 10)

;; 3628800

;; > (john-wallis-pi 10)

;; 3.2137849402931895

;; > (john-wallis-pi 100)

;; 3.1493784731686008

;; > (john-wallis-pi 1000)

;; 3.142377365093878

;; > (john-wallis-pi 10000)

;; 3.1416711865344635

;; > (john-wallis-pi 100000)

;; 3.1416005075027056