学步园

'''
Created on 2012-11-5

@author: Pandara
'''
import math
import random 
import sys

def is_prime(num):
    '''
    determine wether num is prime
    '''
    for i in range(int(math.floor(float(num) / 2)) + 1)[2:]:
        if num % i == 0:
            return False
    
    return True

def random_prime(low = 0, high = 100):
    '''
    to get prime randomly
    low: lower bound of range
    high: higher bound of range
    '''
    
    rand = random.randint(low, high)
    
    while(True):#is prime or not
        if not is_prime(rand):
            rand = (rand + 1) % (high - low) + low
        else:
            return rand

def gcd(a, b):
    '''
    get gcd from a and b
    '''
    if a < b:
        a, b = b, a
        
    while b:
        a, b = b, a % b
    return a

def egcd(a,b):
    '''
    Extended Euclidean Algorithm
    returns x, y, gcd(a,b) such that ax + by = gcd(a,b)
    '''
    u, u1 = 1, 0
    v, v1 = 0, 1
    while b:
        q = a // b
        u, u1 = u1, u - q * u1
        v, v1 = v1, v - q * v1
        a, b = b, a - q * b
    return u, v, a

def modInverse(e,n):
    '''
    d such that de = 1 (mod n)
    e must be coprime to n
    this is assumed to be true
    '''
    return egcd(e,n)[0]%n

def generate_key(e):
    p = random_prime(1000, 2000) #get p and q, to make p and q big enough
    q = random_prime(1000, 2000) #such that e < euler_n
    
    n = p * q #n equal to p * q
    
    euler_n = (p - 1) * (q - 1) #euler(n) = (p - 1)(q - 1)
    
    #adjusting e
    while gcd(euler_n, e) != 1:
        e += 2 #keep it odd
        
    d = modInverse(e, euler_n)
    return e, d, n

def rsa_encrypt(e, n, msg):
    '''
    e: such that ed mod(n) = 1, public, chosen
    n: equal to p * q
    msg: needs encrypting
    return: list with dicimal digits, from elements in msg after encrypted
    '''
    cmsg = []
    
    for elem in msg:
        cmsg.append(pow(ord(elem), e, n))
    
    return cmsg

def rsa_dencrypt(d, n, cmsg):
    '''
    d: such that ed mod(n) = 1, private, calculated
    n: equal to p * q
    cmsg: list with dicimal digits, from elements in msg after encrypted
    '''
    msg = []
    
    for elem in cmsg:
        msg.append("%c" % pow(elem, d, n))
        
    return "".join(msg)

if __name__ == "__main__":
    while(True):
        print "##############menu###############"
        print "1.generate keys"
        print "2.encryption"
        print "3.decryption"
        print "other for quit"
        chose = input("input your choice:")
        
        if chose == 1:
            e, d, n = generate_key(65537)
            print "generate keys as follow:"
            print "e = %d, d = %d, n = %d" % (e, d, n)
            print "public key(e, n) = (%d, %d), private key(%d, %d)" % (e, n, d, n)
        elif chose == 2:
            msg = raw_input("input your message:")
            e, n = input("input the public key(with format \"e,n\"):")
            
            sys.stdout.write("ciphertext:")
            for elem in rsa_encrypt(e, n, msg):
                sys.stdout.write(" %d" % elem)
            sys.stdout.write("\n")
        elif chose == 3:
            cmsg_str = raw_input("input your ciphertext:")
            d, n = input("input the private key(with format \"d,n\"):")
            
            cmsg = []
            for elem in cmsg_str.split(" "):
                cmsg.append(long(elem))
            
            print "plaintext:", rsa_dencrypt(d, n, cmsg)
        else:
            break

尚未解决的问题：如何利用扩展欧几里得算法求模逆元？即egcd及modInverse的实现