Python模拟RSA算法

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参考http://www.liaoxuefeng.com/article/00137389255913089a316bc2ccb48d3b2323759fecd4cf8000

维基百科给出的RSA算法简介如下：假设Alice想要通过一个不可靠的媒体接收Bob的一条私人讯息。她可以用以下的方式来产生一个公钥和一个私钥：    随意选择两个大的质数p和q，p不等于q，计算N=pq。    根据欧拉函数，不大于N且与N互质的整数个数为(p-1)(q-1)    选择一个整数e与(p-1)(q-1)互质，并且e小于(p-1)(q-1)    用以下这个公式计算d：d × e ≡ 1 (mod (p-1)(q-1))    将p和q的记录销毁。(N,e)是公钥，(N,d)是私钥。(N,d)是秘密的。Alice将她的公钥(N,e)传给Bob，而将她的私钥(N,d)藏起来。

#!/usr/bin/env pythondef range_prime(start, end):    l = list()    for i in range(start, end+1):        flag = True        for j in range(2, i):            if i % j == 0:                flag = False                break        if flag:            l.append(i)    return ldef generate_keys(p, q):    #numbers = (11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47)    numbers =range_prime(10, 100)    N = p * q    C = (p-1) * (q-1)    e = 0    for n in numbers:        if n < C and C % n > 0:            e = n            break    if e==0:        raise StandardError("e not found")    d = 0    for n in range(2, C):        if(e * n) % C == 1:            d = n            break    if d==0:        raise StandardError("d not found")    return ((N, e), (N, d))def encrypt(m, key):    C, x = key    return (m ** x) % Cdecrypt = encryptif __name__ == '__main__':    pub, pri = generate_keys(47, 79)    L = range(20, 30)    C = map(lambda x: encrypt(x, pub), L)    D = map(lambda x: decrypt(x, pri), C)    print "keys:", pub, pri    print "message:", L    print "encrypt:", C    print "decrypt:", D

keys: (3713, 11) (3713, 1631)message: [20, 21, 22, 23, 24, 25, 26, 27, 28, 29]encrypt: [406, 3622, 3168, 134, 3532, 263, 1313, 2743, 2603, 1025]decrypt: [20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L]

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