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.