【SICP练习】125 练习3.56

练习3-56

原文

Exercise 3.56. A famous problem, first raised by R. Hamming, is to enumerate, in ascending order with no repetitions, all positive integers with no prime factors other than 2, 3, or 5. One obvious way to do this is to simply test each integer in turn to see whether it has any factors other than 2, 3, and 5. But this is very inefficient, since, as the integers get larger, fewer and fewer of them fit the requirement. As an alternative, let us call the required stream of numbers S and notice the following facts about it.
● S begins with 1.
● The elements of (scale-stream S 2) are also elements of S.
● The same is true for (scale-stream S 3) and (scale-stream 5 S).
● These are all the elements of S.
Now all we have to do is combine elements from these sources. For this we define a procedure merge that combines two ordered streams into one ordered result stream, eliminating repetitions:

(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)))))))))

Then the required stream may be constructed with merge, as follows:

(define S (cons-stream 1 (merge <??> <??>)))

Fill in the missing expressions in the places marked

代码

(define s (cons-stream 1 (merge (scale-stream s 2) (merge (scale-stream s 3) (scale-stream s 5)))))

感谢访问，希望对您有所帮助。 欢迎关注或收藏、评论或点赞。

为使本文得到斧正和提问，转载请注明出处：
http://blog.csdn.net/nomasp

版权声明：本文为 NoMasp柯于旺 原创文章，未经许可严禁转载！欢迎访问我的博客：http://blog.csdn.net/nomasp