Question

您好我正在尝试创建一个正常工作的RSA程序，但在很小的层面上，我在使用此代码进行加密和解密时遇到问题，有人可以帮我弄清楚出了什么问题吗？我尝试过这么多种方式，但这种方式似乎是正确的数学，所以我相信这可能只是我缺乏编码技巧？感谢

import random, math

def RandomPrime():
  prime = False
  while prime == False:
    n = 2
    while n % 2 == 0:
      n = random.randint(10000, 100000)
    s = math.trunc(n**0.5)
    s = int(s)
    x = 3
    # While n doesn't exactly divide to equal 0, and x is less then the sqrt of n
    while ( n % x != 0 ) and (x <= s):
      x = x + 2
    # if n is greater than s, it means it has run out of numbers to test, so is prime
    if x > s:
      prime = True




  return n

def Modulus(p, q):
    M = p * q
    return M

def Totient(p, q):
    T = ((p-1) * (q-1))
    return T

def Pubkey(T):
  prime = False
  while prime == False:
    n = 2
    while n % 2 == 0:
      n = random.randint(3, T)
    s = math.trunc(n**0.5)
    s = int(s)
    x = 3
    # While 
    while ( n % x != 0 ) and (x <= s):
      x = x + 2
    if x > s:
      prime = True
  return n

def privkey( T, n):
    y = math.fmod(1, T)
    d = float((y / n))
    return d


# z is my encyption in this scenario 
z = 8
# I generate p and q, using my random prime generator, i used low primes in
# this example just to see if it would work but it is still not showing reults
p = RandomPrime()
q = RandomPrime()
print(p, q)
#This creates the modulus
M = Modulus(p, q)
print(M)
# Eulier's totient
T = Totient(p, q)
print(T)
#Pub key creation
n =  Pubkey(T)
print(n)
#Priv key creation
d = privkey(n, T)
print(d)


enc = (pow(z, n)) % M
print('enc: ', enc)


dec = (pow(enc, d)) % M
print('dec: ', dec)

Answer 1

您的privkey函数显示错误 - 我猜你看到了RSA私钥值的定义如下：

the value "e" such that e * d = 1 mod Phi(N)

但是在这种情况下，1 mod Phi(N) 不意味着The remainder when 1 is divided by Phi(N)（这似乎是您将其转换为代码的方式，基于您对{{1}的使用但实际上应该更像是：

the value "e" such that (e * d) mod Phi(N) = 1

此值通常使用Extended Euclidean Algorithm计算。 Python实现的示例是here。

值得注意的是，您似乎在定义math.fmod(1, T)，但将其称为privkey(T, n)。

Answer 2

使用python检查我的博客详细包含以下内容的实现：

MD5安全散列算法RFC 1321，RSA公钥加密RFC 3447，OpenPGP RFC 4880

def keyGen():
    ''' Generate  Keypair '''
    i_p=randint(0,20)
    i_q=randint(0,20)
    # Instead of Asking the user for the prime Number which in case is not feasible,
    # generate two numbers which is much highly secure as it chooses higher primes
    while i_p==i_q:
        continue
    primes=PrimeGen(100)
    p=primes[i_p]
    q=primes[i_q]
    #computing n=p*q as a part of the RSA Algorithm
    n=p*q
    #Computing lamda(n), the Carmichael's totient Function.
    # In this case, the totient function is the LCM(lamda(p),lamda(q))=lamda(p-1,q-1)
    # On the Contrary We can also apply the Euler's totient's Function phi(n)
    #  which sometimes may result larger than expected
    lamda_n=int(lcm(p-1,q-1))
    e=randint(1,lamda_n)
    #checking the Following : whether e and lamda(n) are co-prime
    while math.gcd(e,lamda_n)!=1:
        e=randint(1,lamda_n)
    #Determine the modular Multiplicative Inverse
    d=modinv(e,lamda_n)
    #return the Key Pairs
    # Public Key pair : (e,n), private key pair:(d,n)
    return ((e,n),(d,n))

博客链接：Python Cryptography

Github链接：Python Cryptography

简单的RSA代码

2 个答案: