# Hamming's problem of number generation

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-streams 2)` are also elements of `S` .

• The same is true for `(scale-stream S 3)` and `(scale-stream S 5)` .

• 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 `<??>` above.

##### Authentication required

``````(define (stream-car stream) (car stream))

(define (stream-cdr stream) (force (cdr stream)))

(define (stream-ref s n)
(if (= n 0)
(stream-car s)
(stream-ref (stream-cdr s) (- n 1))))

(define (stream-map proc . list-of-stream)
(if (null? (car list-of-stream))
'()
(cons-stream
(apply proc
(map (lambda (s)
(stream-car s))
list-of-stream))
(apply stream-map
(cons proc (map (lambda (s)
(stream-cdr s))
list-of-stream))))))

(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)))))))))

(check-equal? (stream-ref s 0) 1)
(check-equal? (stream-ref s 1) 2)
(check-equal? (stream-ref s 2) 3)
(check-equal? (stream-ref s 3) 4)
(check-equal? (stream-ref s 4) 5)
(check-equal? (stream-ref s 5) 6)
(check-equal? (stream-ref s 6) 8)
(check-equal? (stream-ref s 7) 9)
(check-equal? (stream-ref s 8) 10)
(check-equal? (stream-ref s 9) 12)``````