RSA算法python实现

RSA算法是一种非对称加密算法,是现在广泛使用的公钥加密算法,主要应用是加密信息和数字签名。详情请看维基:http://zh.wikipedia.org/wiki/RSA%E5%8A%A0%E5%AF%86%E6%BC%94%E7%AE%97%E6%B3%95

算法基本思路:

1.公钥与私钥的生成:

(1)随机挑选两个大质数 p 和 q,构造N = p*q;

(2)计算欧拉函数φ(N) = (p-1) * (q-1);

(3)随机挑选e,使得gcd(e, φ(N)) = 1,即 e 与 φ(N) 互素;

(4)计算d,使得 e*d ≡ 1 (mod φ(N)),即d 是e 的乘法逆元。

此时,公钥为(e, N),私钥为(d, N),公钥公开,私钥自己保管。

2.加密信息:

(1)待加密信息(明文)为 M,M < N;(因为要做模运算,若M大于N,则后面的运算不会成立,因此当信息比N要大时,应该分块加密)

(2)密文C = Me mod N

(3)解密Cd mod N = (Me)d mod N = Md*e mod N ;

要理解为什么能解密?要用到欧拉定理(其实是费马小定理的推广)aφ(n) ≡ 1 (mod n),再推广:aφ(n)*k ≡ 1 (mod n),得:aφ(n)*k+1 ≡ a (mod n)

注意到 e*d ≡ 1 mod φ(N),即:e*d = 1 + k*φ(N)。

因此,Md*e mod N = M1 + k*φ(N) mod N = M

简单来说,别人用我的公钥加密信息发给我,然后我用私钥解密。

3.数字签名:

(1)密文C = Md mod N

(2)解密M = Cmod N = (Md)e mod N = Md*e mod N  = M ;(原理同上)

简单来说,我用自己的密钥加密签名,别人用我的公钥解密可以看到这是我的签名。注意,这个不具有隐私性,即任何人都可以解密此签名。

 

算法的安全性:基于大整数N难以分解出p和q,构造φ(N);或由N直接构造φ(N)同样难。

 

算法的实现:

1.快速幂取模;http://www.cnblogs.com/7hat/p/3398394.html

2.素性测试;http://www.cnblogs.com/7hat/p/3400831.html

3.扩展欧几里得求乘法逆元和最大公约数;http://www.cnblogs.com/7hat/p/3406494.html

实现代码:

import random

def fastExpMod(b, e, m):
    """
    e = e0*(2^0) + e1*(2^1) + e2*(2^2) + ... + en * (2^n)

    b^e = b^(e0*(2^0) + e1*(2^1) + e2*(2^2) + ... + en * (2^n))
        = b^(e0*(2^0)) * b^(e1*(2^1)) * b^(e2*(2^2)) * ... * b^(en*(2^n)) 

    b^e mod m = ((b^(e0*(2^0)) mod m) * (b^(e1*(2^1)) mod m) * (b^(e2*(2^2)) mod m) * ... * (b^(en*(2^n)) mod m) mod m
    """
    result = 1
    while e != 0:
        if (e&1) == 1:
            # ei = 1, then mul
            result = (result * b) % m
        e >>= 1
        # b, b^2, b^4, b^8, ... , b^(2^n)
        b = (b*b) % m
    return result

def primeTest(n):
    q = n - 1
    k = 0
    #Find k, q, satisfied 2^k * q = n - 1
    while q % 2 == 0:
        k += 1;
        q /= 2
    a = random.randint(2, n-2);
    #If a^q mod n= 1, n maybe is a prime number
    if fastExpMod(a, q, n) == 1:
        return "inconclusive"
    #If there exists j satisfy a ^ ((2 ^ j) * q) mod n == n-1, n maybe is a prime number
    for j in range(0, k):
        if fastExpMod(a, (2**j)*q, n) == n - 1:
            return "inconclusive"
    #a is not a prime number
    return "composite"

def findPrime(halfkeyLength):
    while True:
        #Select a random number n 
        n = random.randint(0, 1<<halfkeyLength)
        if n % 2 != 0:
            found = True
            #If n satisfy primeTest 10 times, then n should be a prime number
            for i in range(0, 10):
                if primeTest(n) == "composite":
                    found = False
                    break
            if found:
                return n

def extendedGCD(a, b):
    #a*xi + b*yi = ri
    if b == 0:
        return (1, 0, a)
    #a*x1 + b*y1 = a
    x1 = 1
    y1 = 0
    #a*x2 + b*y2 = b
    x2 = 0
    y2 = 1
    while b != 0:
        q = a / b
        #ri = r(i-2) % r(i-1)
        r = a % b
        a = b
        b = r
        #xi = x(i-2) - q*x(i-1)
        x = x1 - q*x2
        x1 = x2
        x2 = x
        #yi = y(i-2) - q*y(i-1)
        y = y1 - q*y2
        y1 = y2
        y2 = y
    return(x1, y1, a)

def selectE(fn, halfkeyLength):
    while True:
        #e and fn are relatively prime
        e = random.randint(0, 1<<halfkeyLength)
        (x, y, r) = extendedGCD(e, fn)
        if r == 1:
            return e

def computeD(fn, e):
    (x, y, r) = extendedGCD(fn, e)
    #y maybe < 0, so convert it 
    if y < 0:
        return fn + y
    return y

def keyGeneration(keyLength):
    #generate public key and private key
    p = findPrime(keyLength/2)
    q = findPrime(keyLength/2)
    n = p * q
    fn = (p-1) * (q-1)
    e = selectE(fn, keyLength/2)
    d = computeD(fn, e)
    return (n, e, d)

def encryption(M, e, n):
    #RSA C = M^e mod n
    return fastExpMod(M, e, n)

def decryption(C, d, n):
    #RSA M = C^d mod n
    return fastExpMod(C, d, n)


#Unit Testing
(n, e, d) = keyGeneration(1024)
#AES keyLength = 256
X = random.randint(0, 1<<256)
C = encryption(X, e, n)
M = decryption(C, d, n)
print "PlainText:", X
print "Encryption of plainText:", C
print "Decryption of cipherText:", M
print "The algorithm is correct:", X == M
Python

 

posted @ 2013-11-05 09:19  7hat  阅读(14423)  评论(6编辑  收藏  举报