SICP Exercise 1.24
This time prime? procedure has been replaced by fast-prime?.
(define (divides? a b)
(= (remainder b a) 0))
(define (square x) (* x x))
(define (expmod base exp m)
(cond ((= exp 0)1)
((even? exp)
(remainder (square (expmod base (/ exp 2) m))
m))
(else
(remainder (* base (expmod base (- exp 1) m))
m))))
(define (fermat-test n)
(define (try-it a)
(= (expmod a n n) a))
(try-it (+ 1 (random (- n 1)))))
(define (fast-prime? n times)
(cond ((= times 0) true)
((fermat-test n) (fast-prime? n (- times 1)))
(else false)))
(define (timed-prime-test n)
(start-prime-test n (runtime)))
(define (start-prime-test n start-time)
(if (fast-prime? n 5)
(report-prime (- (runtime) start-time) n)))
(define (report-prime elapsed-time n)
(display n)
(display "***")
(display elapsed-time)
(newline))
(define (search-for-primes a b)
(if (even? a) (search-for-primes (+ a 1) b)
(cond ((< a b) (timed-prime-test a)
(search-for-primes (+ a 2) b)))))
Again, finding the primes:
> (timed-prime-test 1009)
1009***17
> (timed-prime-test 1013)
1013***17
> (timed-prime-test 1019)
1019***18
> (timed-prime-test 10007)
10007***24
> (timed-prime-test 10009)
10009***21
> (timed-prime-test 10037)
10037***24
> (timed-prime-test 100003)
100003***26
> (timed-prime-test 100019)
100019***27
> (timed-prime-test 100043)
100043***28
> (timed-prime-test 1000003)
1000003***29
> (timed-prime-test 1000033)
1000033***29
> (timed-prime-test 1000037)
1000037**31
The Fermat-test has $$\Theta (\log n)$$ order of growth. From the above tests, it’s clear that the time taken for twice as many digits ($$10^3$$ vs $$10^6$$) is roughly twice as much (17t vs 31t. This supports logarithmic growth.