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Python標準ライブラリを使用したプライベート/パブリック暗号化

次のようなコードを作成できるように、私の検索で発見できなかったモジュールはありますか？このようなコードを書きたい理由は重要ではありません。私が求めているのは、公開および秘密のバイトキーを生成し、それらのキーを使用してデータを簡単にエンコードおよびデコードするシンプルなAPIを備えたコードです。

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

利用できるように見えるもののほとんどは、パッケージをダウンロードする必要があり、Python 2.xでのみ実行されます。PEMファイルまたは他のタイプの証明書で動作するライブラリを見つけることも非常に一般的です。そのようなファイルを処理する必要をなくし、その場で公開鍵と秘密鍵を生成し、メモリ内のデータをすばやく処理するのが好きです。

pythonpython-3.xencryption-asymmetric
18
Noctis Skytower

公開鍵暗号化は標準ライブラリにはありません。 PyPi にはサードパーティのライブラリがいくつかありますが：

その背後にある数学に興味がある場合は、Pythonを使用すると簡単に実験できます。

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

鍵の生成はもう少し複雑です。これは、エントロピーのソースとしてurandomを使用してメモリ内でキーを生成する方法の簡単な例です。コードはPy2.6とPy3.xの両方で実行されます。

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)
32
Raymond Hettinger

これが別の例です

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}
2
Jason

PyCrypto 2.4.1の時点でPython 3で動作します。

2
nmichaels