Question

我正在编写一些RSA实现来帮助我在大学上课，我遇到了一个我无法修复的错误。

这是我到目前为止的代码：

import numpy

def primesfrom2to(n):
    """ Input n>=6, Returns a array of primes, 2 <= p < n 
        Taken from: http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188 """
    sieve = numpy.ones(n/3 + (n%6==2), dtype=numpy.bool)
    for i in xrange(1,int(n**0.5)/3+1):
        if sieve[i]:
            k=3*i+1|1
            sieve[       k*k/3     ::2*k] = False
            sieve[k*(k-2*(i&1)+4)/3::2*k] = False
    return numpy.r_[2,3,((3*numpy.nonzero(sieve)[0][1:]+1)|1)]

def extended_gcd(a, b):
    if b == 0:
        return (1, 0)
    else:
        q = a / b
        r = a - b * q
        s, t = extended_gcd(b, r)
        return (t, s - q * t)

def pick_e(phi):
    primes = primesfrom2to(phi)
    e = primes[0]
    i = 0
    while phi % e != 1 and 1 < e < phi:
        i += 1
        e = primes[i]
    return e

def RSA(p, q, e=None):
    n = p * q
    phi = (p-1) * (q-1)
    if not e:
        e = pick_e(phi)
    x, y = extended_gcd(e, phi)
    d = x % phi
    return (e, n), (d, n)

def encrypt(P, public_key):
    C = P**public_key[0] % public_key[1]
    return C

def decrypt(C, private_key):
    P = C**private_key[0] % private_key[1]
    return P

public_key, private_key = RSA(11, 13)
public_key2, private_key2 = RSA(11, 13, 7)
print public_key, private_key
print public_key2, private_key2

print "public_key and private_key"
print "plaintext -> ciphertext -> plaintext"
for i in range(1,10):
    c =  encrypt(i, public_key)
    p = decrypt(c, private_key)
    print "%9d -> %10d -> %9d" % (i, c, p)

print "public_key2 and private_key2"
print "plaintext -> ciphertext -> plaintext"
for i in range(1,10):
    c =  encrypt(i, public_key2)
    p = decrypt(c, private_key2)
    print "%9d -> %10d -> %9d" % (i, c, p)

我得到的输出是：

(7, 143) (103, 143)
(7, 143) (103, 143)
public_key and private_key
plaintext -> ciphertext -> plaintext
        1 ->          1 ->         1
        2 ->        128 ->         0
        3 ->         42 ->         0
        4 ->         82 ->         0
        5 ->         47 ->        15
        6 ->         85 ->       126
        7 ->          6 ->         0
        8 ->         57 ->        47
        9 ->         48 ->         0
public_key2 and private_key2
plaintext -> ciphertext -> plaintext
        1 ->          1 ->         1
        2 ->        128 ->         2
        3 ->         42 ->         3
        4 ->         82 ->         4
        5 ->         47 ->         5
        6 ->         85 ->         6
        7 ->          6 ->         7
        8 ->         57 ->         8
        9 ->         48 ->         9

两者的输出应该相同，因为键是相同的。谁能看到我做错了什么？

Answer 1

你的pick_e函数显然返回了一个numpy.int64而不是int;如果用

替换return语句

return int(e)

您的代码按预期工作（至少对我而言:-)）

python RSA中相同键的解密输出不同

1 个答案: