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解密密码块链异或“加密”

reverse function python

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Richard Dabel · 技术社区 · 7 年前

我收到了朋友的挑战。但是,我花了几个小时思考逻辑以及如何在代码中写出我的想法。

def hash_cbc(input):
    initial = 'a'
    key = 'netsos'

    output = [ord(initial)]

    def enc(c, key, i):
        z = key[i % len(key)]
        return (c + ord(z)) ^ ord(z)

    for i in range(len(input)):
        prev_char = output[i]
        next_char = enc(prev_char, key, i) ^ ord(input[i])
        output.append(next_char)

    return output

该算法的输出必须为:

97, 194, 299, 411, 533, 654, 768, 873, 965, 1066, 1164, 1272, 1399, 1512, 1603, 1706, 1792, 1895, 1994, 2110, 2215, 2310, 2404, 2517, 2645, 2762

我需要反转函数,以便获得输入。输入必须是字符串。

请帮帮我,我已经被困了几个小时,没有结果。我想反转异或函数,但真的很困惑如何编写它。

我在想这些算法是怎么做的

enc(prev_char, key, i) ^ 1st ordinal input = 1st output
enc(prev_char, key, i) ^ 2nd ordinal input = 2nd output
enc(prev_char, key, i) ^ 3rd ordinal input = 3rd output

但我不知道如何扭转,因为有

我还试着制作自己的反向代码

def enc(c, key, i):
Awalan = input("Insert numbers: ")
Awalan_split = Awalan.split (', ')
Awalan_len = len(Awalan_split)

initial = 'a'
key = 'netsos'
output = [ord(initial)]
z = key[i % len(key)]
return (c + ord(z)) ^ ord(z)

for i in range(Awalan_len):
    prev_char = output[i]
    next_char = int(Awalan_split[i]) ^ enc(prev_char, key, i)
    char_convert = chr(next_char)
    return(char_convert)

print(char_convert)

1 回复 | 直到 7 年前

Mark Tolonen 7 年前

异或反转自身。A^B=C,C^B=A。示例:

>>> 123 ^ 456
435
>>> 435 ^ 456
123

enc() 函数,所以用函数应产生原始值。记住这一点 initial 是 'a' enc 0 :

>>> 194 ^ enc(ord('a'),'netsos',0)
99
>>> chr(99)
'c'

结果的第三个值299使用序列的前一个值和下一个字符与加密函数异或,因此再次与加密函数异或应产生它:

>>> 299 ^ enc(194,key,1) # index (last parm) incremented
105
>>> chr(105)
'i'

output = []
for i in range(len(inp) - 1): # original will be one shorter.
    prev_char = inp[i]
    next_char = enc(prev_char, key, i) ^ inp[i+1]
    output.append(next_char)

然后只需将生成的序数转换回字符:

return ''.join(chr(n) for n in output)

以下是所有联系在一起的内容:

# factor out common stuff
initial = 'a'
key = 'netsos'
def enc(c, key, i):
    z = key[i % len(key)]
    return (c + ord(z)) ^ ord(z)

def hash_cbc(inp):
    output = [ord(initial)]

    for i in range(len(inp)):
        prev_char = output[i]
        next_char = enc(prev_char, key, i) ^ ord(inp[i])
        output.append(next_char)

    return output

def unhash_cbc(inp):
    output = []
    for i in range(len(inp) - 1):
        prev_char = inp[i]
        next_char = enc(prev_char, key, i) ^ inp[i+1]
        output.append(next_char)

    return ''.join(chr(n) for n in output)

expected_result = [97, 194, 299, 411, 533, 654, 768, 873, 965, 1066, 1164, 1272, 1399, 1512, 1603, 1706, 1792, 1895, 1994, 2110, 2215, 2310, 2404, 2517, 2645, 2762]

u = unhash_cbc(expected_result)
h = hash_cbc(u)
print(h == expected_result)
print(u)

True
cipheringblockingchaining