sicp 流 函数(3-53 to 3-58,3-59,3-60,3-61,3-62 答案)
;;; -------- 3.53 ------------------------------
(define (add-stream s1 s2)
(stream-map + s1 s2))
define s (cons-stream 1 (add-stream s s)))
定义 2^n 流函数(n为从0开始的整数， 为流的第n个元素)
;;; --------- 3.54 -----------------------------
(define (mul-streams s1 s2)
(stream-map * s1 s2))
(define (add-streams s1 s2)
(stream-map + s1 s2))
(define ones (cons-stream 1 ones))
(define integers (cons-stream 1 (add-streams ones integers)))
(define factorials (cons-stream 1 (mul-streams factorials integers)))
;;; ----------- 3.55 ----------------------------
(define (partial-sums stream)
(define sumers
(cons-stream (stream-car stream)
(add-streams sumers
(stream-cdr stream))))
sumers)
;;; ------------ 3.56 --------------------
(define (scale-stream stream factor)
(stream-map (lambda (x) (* x factor)) stream))
(define (merge s1 s2)
(cond ((stream-null? s1) s2)
((stream-null? s2) s1)
(else
(let ((s1car (stream-car s1))
(s2car (stream-car s2)))
(cond ((< s1car s2car)
(cons-stream s1car (merge (stream-cdr s1) s2)))
((> s1car s2car)
(cons-stream s2car (merge s1 (stream-cdr s2))))
(else
(cons-stream s1car (merge (stream-cdr s1)
(stream-cdr s2)))))))))
(merge (scale-stream S 3)
(scale-stream S 5)))))
;;; --------------- 3.57 -----------------------------
如果简单地用(lambda () exp) 实现 (delay exp)的话， 每次获取流中的第n个元素，都得从头使用fibgen计算到第n个元素为止，
与正常的迭代计算斐波那契数所需的工作量一般大。 所以带记忆版本的流有助于减少计算的工作量，不过需要更多额外的内存空间。
;;; --------------- 3.58 -----------------------------
(define (expand num den radix)
(cons-stream
(quotient (* num radix) den)
(expand (remainder (* num radix) den) den radix)))
; 这段程序求 num/den， 并以radix为基数构成一个无限流。
(expand 1 7 10) ;结果为 1/7 = (1 4 2 8 5 7 1 4 2 8 5 7 ....)
(expand 3 8 10) ;结果为 3/8 = (3 7 5 0 0 0 0 0 0 0 0 0 ....)
;;; --------------- 3.59 -----------------------------
a. 流的每个元素除以整数就可以
(define (integrate-series stream)
(div-streams stream integers))
b. 上面定义的 (integrate-series stream) 结果可认为是stream流的积分(缺少常数项)。
ex的倒数等于ex，而ex的常数项为1，所以ex也等于 1加上ex的积分
(define exp-series
(cons-stream 1 (integrate-series exp-series)))
由于sin的倒数是cos， 所以sin就是 cos的倒数加上一个常数项， 有 sin的多项式定义可得：
(define sine-series
(cons-stream 0 (integrate-series cosine-series)))
同理获得cos的定义，cos和sin两个级数递归调用形成无限流
(define cosine-series
(cons-stream 1 (stream-map (lambda (x) (- x))
(integrate-series sine-series))))
;;; --------------- 3.60 -----------------------------
两个级数相乘，等于级数s1的每一项与s2中的每一项相乘(注意：不是一对一的相乘)
方法是: s1和s2相乘所得结果的第一项肯定是s1的第一项乘以s2的第一项，
至于第二项后的结果为：s1的第一项与s2的第二项以及之后的项相乘之和， 再加上s1除去第一项与s2的所有项相乘所得结果。
(define (mul-series s1 s2)
(cons-stream (*
(stream-car s1)
(stream-car s2))
(stream-map + (scale-stream
(stream-cdr s2)
(stream-car s1))
(mul-series
(stream-cdr s1)
s2))))
使用sin2x + cos2x = 1验证：
define one (add-streams (mul-series sine-series sine-series)
(mul-series cosine-series cosine-series)))
(stream-ref one 0) ==> 1
(stream-ref one N) ==> 0 ; N表示除了0以外的其他数
;;; --------------- 3.61 -----------------------
;;; 定义常数项为1的级数的倒数
(define (invert-unit-series s)
(define invert-dummy
(cons-stream 1 (scale-stream
(mul-series (stream-cdr s) invert-dummy)
-1)))
invert-dummy)
;;; -------------- 3..62 ---------------------
;;; 先定义第一项非零的级数的倒数
(define (invert-series s)
(let ((constant-term (stream-car s)))
(if (= constant-term 0)
(error "DIV-SERIES denom constant term can't be zero")
(scale-stream (invert-unit-series
(scale-stream s (/ 1 constant-term)))
constant-term))))
;;; 两个级数相除等级分子的级数乘以分母的级数
(define (div-series num denom)
(mul-series num
(invert-series denom)))
;;; 定义tan的级数， tan = sin / cos
(define tan-series (div-series sine-series cosine-series))
;;; 下面显示 tan级数前10项的值
(stream-ref tan-series 0)
1 ]=>
;Value: 0
(stream-ref tan-series 1)
1 ]=>
;Value: 1
(stream-ref tan-series 2)
1 ]=>
;Value: 0
(stream-ref tan-series 3)
1 ]=>
;Value: 1/3
(stream-ref tan-series 4)
1 ]=>
;Value: 0
(stream-ref tan-series 5)
1 ]=>
;Value: 2/15
(stream-ref tan-series 6)
1 ]=>
;Value: 0
(stream-ref tan-series 7)
1 ]=>
;Value: 17/315
(stream-ref tan-series 8)
1 ]=>
;Value: 0
(stream-ref tan-series 9)
1 ]=>
;Value: 62/2835
1 ]=>
参考： http://wizardbook.wordpress.com/solutions-index/ ------- SICP参考答案