Encrypter - Python中的双Vigenere密码

Question

我根据Vigenere密码的思想编写了这个加密器，但是它不是只使用一个密钥，而是从现有密钥中“生成”另一个密钥。第二个键的长度也取决于第一个键中的不同字符。因此，它在转换字母时使用了这两个键。

def scram(key):
    key2 = []
    #make the length of the second key varying from key to key for harder cracking
    length = (key[0]-32)+(key[int(len(key)/2)]-32)+(key[len(key)-1]-32)
    #make the max length = 256
    length = length % 256
    #if the length is less than 64, multiply by two to make it longer
    while(length < 64):
        length*=2

    #scrambles the letters around
    for x in range(length):
        #basically shifts the key according to the current character,
        #how many times it has looped the key, where it is in the key,
        #and the modulo of x and a few prime numbers to make sure that 
        #an overlap/repeat doesn't happen.
        toapp = (x%(len(key)-1)) + (x/(len(key)-1)) + (key[x%(len(key)-1)]) + (x%3) +(x%5)+(x%53)+(x%7)                   
        toapp = int(toapp % 94)
        key2.append(toapp)

    return key2

def cipher(mes,key,ac):
    #makes the second key
    key2 = scram(key)
    res=[]

    #do proper shifting in the keys
    for x in range(len(mes)):
        temp=mes[x]

        if(action == "2"):
            temp = temp - key[x%(len(key)-1)] - key2[x%(len(key2)-1)]
        else:
            temp = temp + key[x%(len(key)-1)] + key2[x%(len(key2)-1)]                

        temp = int(temp % 94)

        res.append(chr(temp+32))

    return res

#encrypt or decrypt
action = input("type 1 to encypt. type 2 to decrypt:")
#input
m= input("Text:")
k= input("key - 4 or more char:")

#changes the letters to ascii value
mes= []
for x in m:
    mes.append(ord(x)-32)

key= []
for x in k:
    key.append(ord(x)-32)

#encrypts it    
result = cipher(mes,key,action)

for x in result:
    print(x,end="")

print("")

y = input("Press enter to continue...")

有没有更有效的方法来做到这一点？这是加密文本的安全方法吗？你能破解这个程序加密的文本吗？

Answer 1

列表理解

m= input("Text:")
k= input("key - 4 or more char:")

#changes the letters to ascii value
mes= []
for x in m:
    mes.append(ord(x)-32)

key= []
for x in k:
    key.append(ord(x)-32)

这一切都可以缩短为两个清单的理解：

mes = [ord(x) - 32 for x in input("Text: ")]
key = # Exercise for the reader

避免函数

中的全局

    if(action == "2"):

动作应该是函数的一个参数。

temp永存

 temp=mes[x]

您命名一个变量temp并使用它8行..。请给这个重要的变量起一个合理的名字。

超长公式

toapp = (x%(len(key)-1)) + (x/(len(key)-1)) + (key[x%(len(key)-1)]) + (x%3) +(x%5)+(x%53)+(x%7)                   

请评论一下这个巨大的公式。

避免了这么多作业，

result = cipher(mes,key)

for x in result:
    print(x,end="")

print("")

在python中，避免外部变量赋值是惯用的，而显式循环很少是真正必要的，这一切都可以缩短为：

print(''.join(cipher(mes,key)))

不分配垃圾输入

最后一行应该是：

input("Enter to continue...")