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(define (smallest-divisor n) (find-divisor n 2)) (define (find-divisor n test-divisor) (cond ((> (square test-divisor) n) n) ((divides? test-divisor n) test-divisor) (else (find-divisor n (next test-divisor))))) (define (next test-divisor) (if (= test-divisor 2) 3 (+ test-divisor 2))) (define (divides? a b) (= (remainder b a ) 0)) (define (prime? n) (= n (smallest-divisor n))) (define (square n) (* n n)) (define (sfp count n) (define (timed-prime-test n) (start-prime-test n (current-process-milliseconds))) (define (start-prime-test n start-time) (if (prime? n) (report-prime (- (current-process-milliseconds) start-time)) (sfp count (+ n 1)))) (define (report-prime elapsed-time) (display " *** ") (display elapsed-time) (sfp (- count 1) (+ n 1))) ; sfp ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; (if (= count 0) (display "\ndone") (timed-prime-test n))) ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; (sfp 10 (* 1000 1000)) ; .33333 (sfp 6 (* 10000 1000)) ; 1 (sfp 6 (* 100000 1000)) ; 4 (sfp 6 (* 1000000 1000)) ; 13 (sfp 6 (* 10000000 1000)) ; 84 (sfp 6 (* 100000000 1000)) ; 280 ; It does seem to cut the amount of time in half. The questions in the exercise ; make it seem like it shouldn't be this way though.

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