我一直在努力寻找编码数和解码数。我知道你需要找到这些的基本规则,并开始尝试创建一个 python 程序,它从前 150 个素数的随机选择中找到 e 变量和 d 变量。我这样做的方式是:
import random
from primesieve import *
from sympy.solvers import *
from sympy import *
f = 0
choiceOfPrimes = generate_primes(10)
p = random.choice(choiceOfPrimes)
q = random.choice(choiceOfPrimes)
while p == q:
p = random.choice(choiceOfPrimes)
n = p * q
phiN = (p-1) * (q-1)
lista = []
listb = []
for naa in range(1, phiN):
lista.append(naa)
for naaa in range(1, 35):
listb.append(naaa)
e = random.choice(lista)
def finding_d():
global phiN
global e
global f
f = e
abc = []
abc = divmod(phiN, f)
abc1 = f * abc[0]
abc2 = abc[1]
phiN = f * abc1 + abc2 # phiN = f * {how many it goes in} + {remainder}
abc3 = divmod(f, abc2)
f = abc2 * abc3[0] + abc3[1] # (moved f to phiN) and (remainder to f)
abc4 = divmod(abc2, abc3[1])
e1 = abc3[1] * abc4[0]
e2 = abc4[1]
abc2 = e1 + e2
e3 = abc2 + (e1*-1) # This bit I am struggling
d = f
while (e % n) == 0 and (e % phiN) == 0:
for r in range(phiN, 0, -1):
e = r
while ((d * e) % phiN) != 1:
for r in range(1, 35):
e = r
print(p, q, n, phiN, e, d)
它需要永远运行并且从未完成运行。我什至尝试将 generate_primes(150) 更改为 generate_primes(10) 但同样的问题发生了。
有没有人有解决方案,我很乐意听到(顺便说一下,primesieve 库不会自动包含在 python 库中,您必须自己下载)。谢谢。
编辑:
我做了: 而 p == q: p = random.choice(choiceOfPrimes) 但是我很难做欧几里德垫子
最佳答案
你的代码有很多问题:
首先,
while p == q:
p = random.choice(choiceOfPrimes)
You should do this step before calculating the value of
phiN
as the value of phiN would change if you change the value ofp
.
其次,
d = random.choice(lista)
For calculating
d
you should find the Modular multiplicative inverse ofe
with respect tophiN
using theExtended Euclidean algorithm
, choosing values ofe
untill they work is very inefficient.
关于python - RSA 编码和解码 [e,d]。在 python 中查找 e 和 d.. 更新,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36814887/