如何在python中找到pow(a,b,c)的反向? [英] How to find reverse of pow(a,b,c) in python?
问题描述
pow(a,b,c)
运算符返回(a**b)%c
.如果我有b
,c
的值以及此操作(res=pow(a,b,c))
的结果,如何找到a
的值?
pow(a,b,c)
operator in python returns (a**b)%c
. If I have values of b
, c
, and the result of this operation (res=pow(a,b,c))
, how can I find the value of a
?
推荐答案
尽管注释中的语句这不是离散对数问题,但不是.这更类似于 RSA问题,其中c
是两个结果的乘积大素数,b
是加密指数,a
是未知的明文.我总是喜欢使x
为您要解决的未知变量,因此您具有y
= x b mod c,其中已知y
,b
和c
,您想求解x
.解决它涉及与RSA中相同的基本数论,即必须计算z
= b -1 modλ(c),然后可以通过x
=来求解x
. y z mod c. λ是 Carmichael的lambda函数,但是您也可以改用Euler的phi(totient)函数.我们已经将原始问题简化为计算逆模λ(c).如果c
易于分解,或者我们已经知道c
的分解,那么就很容易做到,否则很难.如果c
小,那么蛮力是可以接受的技术,您可以忽略所有复杂的数学运算.
Despite the statements in the comments this is not the discrete logarithm problem. This more closely resembles the RSA problem in which c
is the product of two large primes, b
is the encrypt exponent, and a
is the unknown plaintext. I always like to make x
the unknown variable you want to solve for, so you have y
= xb mod c where y
, b
, and c
are known, you want to solve for x
. Solving it involves the same basic number theory as in RSA, namely you must compute z
=b-1 mod λ(c), and then you can solve for x
via x
= yz mod c. λ is Carmichael's lambda function, but you can also use Euler's phi (totient) function instead. We have reduced the original problem to computing an inverse mod λ(c). This is easy to do if c
is easy to factor or we already know the factorization of c
, and hard otherwise. If c
is small then brute-force is an acceptable technique and you can ignore all the complicated math.
以下是显示这些步骤的代码:
Here is some code showing these steps:
import functools
import math
def egcd(a, b):
"""Extended gcd of a and b. Returns (d, x, y) such that
d = a*x + b*y where d is the greatest common divisor of a and b."""
x0, x1, y0, y1 = 1, 0, 0, 1
while b != 0:
q, a, b = a // b, b, a % b
x0, x1 = x1, x0 - q * x1
y0, y1 = y1, y0 - q * y1
return a, x0, y0
def inverse(a, n):
"""Returns the inverse x of a mod n, i.e. x*a = 1 mod n. Raises a
ZeroDivisionError if gcd(a,n) != 1."""
d, a_inv, n_inv = egcd(a, n)
if d != 1:
raise ZeroDivisionError('{} is not coprime to {}'.format(a, n))
else:
return a_inv % n
def lcm(*x):
"""
Returns the least common multiple of its arguments. At least two arguments must be
supplied.
:param x:
:return:
"""
if not x or len(x) < 2:
raise ValueError("at least two arguments must be supplied to lcm")
lcm_of_2 = lambda x, y: (x * y) // math.gcd(x, y)
return functools.reduce(lcm_of_2, x)
def carmichael_pp(p, e):
phi = pow(p, e - 1) * (p - 1)
if (p % 2 == 1) or (e >= 2):
return phi
else:
return phi // 2
def carmichael_lambda(pp):
"""
pp is a sequence representing the unique prime-power factorization of the
integer whose Carmichael function is to be computed.
:param pp: the prime-power factorization, a sequence of pairs (p,e) where p is prime and e>=1.
:return: Carmichael's function result
"""
return lcm(*[carmichael_pp(p, e) for p, e in pp])
a = 182989423414314437
b = 112388918933488834121
c = 128391911110189182102909037 * 256
y = pow(a, b, c)
lam = carmichael_lambda([(2,8), (128391911110189182102909037, 1)])
z = inverse(b, lam)
x = pow(y, z, c)
print(x)
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