Question

是否有一个模块让我的搜索无法发现允许编写如下代码？想要编写这样的代码的原因并不重要。我所追求的是一些代码，它有一个简单的API来生成公钥和私钥字节密钥，并使用这些密钥轻松编码和解码数据。

import module, os

method, bits, data = 'RSA', 1024, os.urandom(1024)
public, private = module.generate_keys(method, bits)

assert isinstance(public, bytes) and isinstance(private, bytes)
assert module.decode(module.encode(data, private), public) == data
assert module.decode(module.encode(data, public), private) == data

大多数似乎可用的内容都需要下载程序包，并且只能在Python 2.x上运行。查找使用PEM文件或其他类型证书的库也很常见。我想避免处理这些文件，动态生成公钥和私钥，并快速处理内存中的数据。

Answer 1

公钥加密不在标准库中。虽然在PyPi上有一些第三方库：

如果你对它背后的数学感兴趣，Python可以很容易地进行实验：

code = pow(msg, 65537, 5551201688147)               # encode using a public key
plaintext = pow(code, 109182490673, 5551201688147)  # decode using a private key

密钥生成涉及更多一点。下面是如何使用 urandom 作为熵源在内存中进行密钥生成的简化示例。代码在Py2.6和Py3.x下运行：

import random

def gen_prime(N=10**8, bases=range(2,20000)):
    # XXX replace with a more sophisticated algorithm
    p = 1
    while any(pow(base, p-1, p) != 1 for base in bases):
        p = random.SystemRandom().randrange(N)
    return p

def multinv(modulus, value):
    '''Multiplicative inverse in a given modulus

        >>> multinv(191, 138)
        18
        >>> 18 * 138 % 191
        1

    '''
    # http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm
    x, lastx = 0, 1
    a, b = modulus, value
    while b:
        a, q, b = b, a // b, a % b
        x, lastx = lastx - q * x, x
    result = (1 - lastx * modulus) // value
    return result + modulus if result < 0 else result

def keygen(N):
    '''Generate public and private keys from primes up to N.

        >>> pubkey, privkey = keygen(2**64)
        >>> msg = 123456789012345
        >>> coded = pow(msg, 65537, pubkey)
        >>> plain = pow(coded, privkey, pubkey)
        >>> assert msg == plain

    '''
    # http://en.wikipedia.org/wiki/RSA
    prime1 = gen_prime(N)
    prime2 = gen_prime(N)
    totient = (prime1 - 1) * (prime2 - 1)
    return prime1 * prime2, multinv(totient, 65537)

Answer 2

PyCrypto适用于Python 3及2.4.1。

Answer 3

这是另一个例子

import random


# RSA Algorithm



ops = raw_input('Would you like a list of prime numbers to choose from (y/n)? ')
op = ops.upper()

if op == 'Y':
    print """\n 2      3      5      7     11     13     17     19     23     29 
31     37     41     43     47     53     59     61     67     71 
73     79     83     89     97    101    103    107    109    113 
127    131    137    139    149    151    157    163    167    173 
179    181    191    193    197    199    211    223    227    229 
233    239    241    251    257    263    269    271    277    281 
283    293    307    311    313    317    331    337    347    349 
353    359    367    373    379    383    389    397    401    409 
419    421    431    433    439    443    449    457    461    463 
467    479    487    491    499    503    509    521    523    541 
547    557    563    569    571    577    587    593    599 \n"""
    rsa()
else:
    print "\n"
    rsa()

def rsa():
    # Choose two prime numbers p and q
    p = raw_input('Choose a p: ')
    p = int(p)

while isPrime(p) == False:
    print "Please ensure p is prime"
    p = raw_input('Choose a p: ')
    p = int(p)

q = raw_input('Choose a q: ')
q = int(q)

while isPrime(q) == False or p==q:
    print "Please ensure q is prime and NOT the same value as p"
    q = raw_input('Choose a q: ')
    q = int(q)

# Compute n = pq
n = p * q

# Compute the phi of n
phi = (p-1) * (q-1)

# Choose an integer e such that e and phi(n) are coprime
e = random.randrange(1,phi)

# Use Euclid's Algorithm to verify that e and phi(n) are comprime
g = euclid(e,phi)
while(g!=1):
    e = random.randrange(1,phi)
    g = euclid(e,phi)

# Use Extended Euclid's Algorithm 
d = extended_euclid(e,phi)

# Public and Private Key have been generated
public_key=(e,n)
private_key=(d,n)
print "Public Key [E,N]: ", public_key
print "Private Key [D,N]: ", private_key

# Enter plain text to be encrypted using the Public Key
sentence = raw_input('Enter plain text: ')
letters = list(sentence)

cipher = []
num = ""

# Encrypt the plain text
for i in range(0,len(letters)):
    print "Value of ", letters[i], " is ", character[letters[i]]

    c = (character[letters[i]]**e)%n
    cipher += [c]
    num += str(c)
print "Cipher Text is: ", num

plain = []
sentence = ""

# Decrypt the cipher text    
for j in range(0,len(cipher)):

    p = (cipher[j]**d)%n

    for key in character.keys():
        if character[key]==p:
            plain += [key]
            sentence += key
            break
print "Plain Text is: ", sentence

# Euclid's Algorithm
def euclid(a, b):
   if b==0:
   return a
else:
   return euclid(b, a % b)

# Euclid's Extended Algorithm
def extended_euclid(e,phi):
    d=0
    x1=0
    x2=1
    y1=1
    orig_phi = phi
    tempPhi = phi

while (e>0):
  temp1 = int(tempPhi/e)
  temp2 = tempPhi - temp1 * e
  tempPhi = e
  e = temp2

  x = x2- temp1* x1
  y = d - temp1 * y1

  x2 = x1
  x1 = x
  d = y1
  y1 = y

  if tempPhi == 1:
      d += phi
      break
return d

# Checks if n is a prime number
def isPrime(n):
   for i in range(2,n):
    if n%i == 0:
        return False
return True

character = {"A":1,"B":2,"C":3,"D":4,"E":5,"F":6,"G":7,"H":8,"I":9,"J":10,
     "K":11,"L":12,"M":13,"N":14,"O":15,"P":16,"Q":17,"R":18,"S":19,
     "T":20,"U":21,"V":22,"W":23,"X":24,"Y":25,"Z":26,"a":27,"b":28,
     "c":29,"d":30,"e":31,"f":32,"g":33,"h":34,"i":35,"j":36,"k":37,
     "l":38,"m":39,"n":40,"o":41,"p":42,"q":43,"r":44,"s":45,"t":46,
     "u":47,"v":48,"w":49,"x":50,"y":51,"z":52, " ":53, ".":54, ",":55,
     "?":56,"/":57,"!":58,"(":59,")":60,"$":61,":":62,";":63,"'":64,"@":65,
     "#":66,"%":67,"^":68,"&":69,"*":70,"+":71,"-":72,"_":73,"=":74}

Answer 4

这里在很多地方都提到了 PyCrypto。有了这个答案，而不是评论，我想指出 PyCrypto 的最后一个版本是从 2013 年开始的。对于密码学库来说，这是不可接受的，因为潜在的危险问题没有得到解决。

但是，PyCrypto 有一个分支，PyCryptodome，它将继续开发。 Link to Github: https://github.com/Legrandin/pycryptodome

使用标准库在Python中进行私有/公共加密

4 个答案: