Question

Assuming n is a positive integer, the composite function performs as follows:

(define (composite? n)
  (define (iter i)
    (cond ((= i n) #f)
          ((= (remainder n i) 0) #t)
          (else (iter (+ i 1)))))
  (iter 2))

It seems to me that the time complexity (with a tight bound) here is O(n) or rather big theta(n). I am just eyeballing it right now. Because we are adding 1 to the argument of iter every time we loop through, it seems to be O(n). Is there a better explanation?

Answer 1

The function as written is O(n). But if you change the test (= i n) to (< n (* i i)) the time complexity drops to O(sqrt(n)), which is a considerable improvement; if n is a million, the time complexity drops from a million to a thousand. That test works because if n = pq, one of p and q must be less than the square root of n while the other is greater than the square root of n; thus, if no factor is found less than the square root of n, n cannot be composite. Newacct’s answer correctly suggests that the cost of the arithmetic matters if n is large, but the cost of the arithmetic is log log n, not log n as newacct suggests.