[SICP] 問題 1.21 : 最小除数の探索

smallest-divisor手続きを使い, 次の数の最小除数を見つけよ: 199, 1999, 19999.

テキストの手続きを入力します.

(define (square x) (* x x))

(define (divides? a b)
  (= (remainder b a) 0))

(define (find-divisor n test-divisor)
  (cond ((> (square test-divisor) n) n)
        ((divides? test-divisor n) test-divisor)
        (else (find-divisor n (+ test-divisor 1)))))

(define (smallest-divisor n)
  (find-divisor n 2))

(define (prime? n)
  (= n (smallest-divisor n)))

評価してみます.

ようこそ DrRacket, バージョン 6.1 [3m].
言語: Pretty Big; memory limit: 2048 MB.
> (smallest-divisor 199)
199
> (smallest-divisor 1999)
1999
> (smallest-divisor 19999)
7
>