'''Created on 2012-11-5@author: Pandara'''import mathimport random import sysdef 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 Truedef 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 randdef gcd(a, b): ''' get gcd from a and b ''' if a < b: a, b = b, a while b: a, b = b, a % b return adef 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, adef 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]%ndef 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, ndef 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 cmsgdef 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
