SICP 练习1.28

Miller-Rabin检查

k2=pn+1

(k−1)(k+1)=pn

如果n为质数，则n=k−1或者n=k+1

n=k+1即k=n−1

n=k−1即k=n+1这是不可能的，因为k<n

用let定义局部变量的时候出错了，就是首先让a=一个数值，然后让a去定义b,这样是不行，lisp中不允许这样干。于是去网上找了找，发现一个函数版的，于是先用函数实现了一遍，发现可以，然后把自己原先的代码用let实现了。

函数调用

(define (expmod base exp m)  (define check     (lambda (a1)      (let ((a2 (remainder (square a1) m)))              (if (and (= a2 1) (> a1 1) (< a1 (- m 1)))                 0 a2))))   (cond ((= exp 0) 1)        ((even? exp) (check (expmod base (/ exp 2) m)))         (else (remainder (* base (expmod base (- exp 1) m)) m))))(define (Miller-Rabin-test n)  (define (try-it a)    (= (expmod a (- n 1) n) 1)) ; a ^ (n-1) % n == 1   (try-it (+ 1 (random (- n 1)))))(define (fast-prime? n times)  (cond ((= times 0) #t)        ((Miller-Rabin-test n) (fast-prime? n (- times 1)))        (else #f)))(define (even? x)  (= (remainder x 2) 0))(define (square x)  (* x x))(fast-prime? 11 3)

两个let

(define (expmod base exp m)   (cond ((= exp 0) 1)        ((even? exp) (let ((a1 (expmod base (/ exp 2) m)))                        (let ((a2 (remainder (square a1) m)))                         (if (and (= a2 1) (> a1 1) (< a1 (- m 1))) 0 a2))))        (else (remainder (* base (expmod base (- exp 1) m)) m))))(define (Miller-Rabin-test n)  (define (try-it a)    (= (expmod a (- n 1) n) 1)) ; a ^ (n-1) % n == 1   (try-it (+ 1 (random (- n 1)))))(define (fast-prime? n times)  (cond ((= times 0) #t)        ((Miller-Rabin-test n) (fast-prime? n (- times 1)))        (else #f)))(define (even? x)  (= (remainder x 2) 0))(define (square x)  (* x x))(fast-prime? 11 3)