sicp习题试解 (2.38)

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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 08/19/05

; modified:

; (保持内容完整不变前提下，可以任意转载)

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;; SICP No.2.38

;; 原文中的fold-left定义有些繁琐，修改如下（同样是迭代）

(define (fold-left op initial sequence)

(if (null? sequence)

initial

(fold-left op (op initial (car sequence)) (cdr sequence))))

(define fold-right accumulate)

> (fold-right / 1 (list 1 2 3))

3/2

> (fold-left / 1 (list 1 2 3))

1/6

> (fold-right list () (list 1 2 3))

(1 (2 (3 ())))

> (fold-left list () (list 1 2 3))

(((() 1) 2) 3)

;; 易证，当op满足交换律和结合律时

;; 即 (op a b) = (op b a)）

;; (op (op a b) c) = (op a (op b c)) 时

;; fold-left和fold-right可以保证产生同样的结果