1.24¶
; Modify the timed-prime-test procedure of Exercise 1.22 ; to use fast-prime? (the Fermat method), and test each ; of the 12 primes you found in that exercise. Since the ; Fermat test has Θ(log n) growth, how would you expect ; the time to test primes near 1,000,000 to compare with ; the time needed to test primes near 1000? Do your data ; bear this out? Can you explain any discrepancy you find? ; Let's define fast-prime procedure (define (even? x) (= (remainder x 2) 0)) (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 (prime? n) (fast-prime? n 500)) ; Define `timed-prime-test` procedure (define (timed-prime-test n) (newline) (display n) (start-prime-test n (runtime))) (define (start-prime-test n start-time) (if (prime? n) (report-prime (- (runtime) start-time)))) (define (report-prime elapsed-time) (display " *** ") (display elapsed-time)) ; Test the timing on large prime numbers (timed-prime-test 1000000007) ; .05000000000000002 (timed-prime-test 1000000009) ; .03999999999999998 (timed-prime-test 1000000021) ; .04000000000000001 (timed-prime-test 10000000019) ; .06 (timed-prime-test 10000000033) ; .04999999999999999 (timed-prime-test 10000000061) ; .06 (timed-prime-test 100000000003) ; .06 (timed-prime-test 100000000019) ; .06000000000000005 (timed-prime-test 100000000057) ; .06000000000000005 (timed-prime-test 1000000000039) ; .06999999999999995 (timed-prime-test 1000000000061) ; .06000000000000005 ; Test by doubling number of digits (timed-prime-test 1000003) ; 3.9999999999999925e-2 (timed-prime-test 1000000000063) ; .08000000000000007 ; i.e. ; 10^6 => 0.04 ; 10^12 => 0.08, which is a logarithmic scale