Question

我在试图找出如何让我的解密功能工作时遇到了很多麻烦。它适用于msg可以被3整除的理想情况，但之后我完全迷失了。我对我必须做的事情有一个模糊的想法，因为我能够使用两个轨道来实现这一点，但是有三个轨道的可能性更大。我很丢失:(所有那些印刷声明只是试图帮我弄清楚发生了什么。

import sys

def main():
    plaintext="abcdefgh"
    print(threeRailEncrypt(plaintext))
    print(threeRailDecrypt(threeRailEncrypt(plaintext)))


def threeRailEncrypt(plaintext):
    ciphertext=""
    rail1=""
    rail2=""
    rail3=""

    for i in range(len(plaintext)):
        if i%3 == 0:
            rail1=rail1+plaintext[i]
        elif i%3 == 1:
            rail2=rail2+plaintext[i]
        else:
            rail3=rail3+plaintext[i]

    ciphertext=rail1+rail2+rail3

    return(ciphertext)

def threeRailDecrypt(msg):
    if len(msg)%3==0:
        third=len(msg)//3
        print(third)
        rail1=msg[:third]
        rail2=msg[third:third*2]
        rail3=msg[third*2:]
        print(rail1,rail2,rail3)
        dm=""
        for i in range(third):
            dm=dm+rail1[i]
            dm=dm+rail2[i]
            dm=dm+rail3[i]
    else:
        third=(len(msg)//3)+1
        print(third)
        rail1=msg[:third]
        rail2=msg[third:third*2]
        rail3=msg[third*2:]
        print(rail1,rail2,rail3)
        dm=""
        for i in range(third):
            dm=dm+rail1[i]
            print(dm)
            dm=dm+rail2[i]
            print(dm)
            dm=dm+rail3[i]
            print(dm)
            if  len(rail2)>len(rail3):
                dm=dm+rail2[-1]
        return(dm)
main()

Answer 1

您可以使用步幅切割字符串（或任何其他序列）;您可以选择每个第3个字符：

def threeRailEncrypt(plaintext):
    return plaintext[::3] + plaintext[1::3] + plaintext[2::3]

要扭转趋势，请将密文分成三个块并使用itertools.zip_longest()重新组合它们：

from itertools import zip_longest, chain

def threeRailDecrypt(ciphertext):
    stride, remainder = divmod(len(ciphertext), 3)
    # how large was each of the three sections?
    sizes = [stride] * 3
    for i in range(remainder):
        sizes[i] += 1
    # slice ciphertext up into 3 sections again
    pos = 0
    pieces = []
    for s in sizes:
        pieces.append(ciphertext[pos:pos + s])
        pos += s
    # recombine the triplets
    return ''.join(chain.from_iterable(zip_longest(*pieces, fillvalue='')))

演示：

>>> threeRailEncrypt('Foo bar baz and all')
'F raa lob znaoab dl'
>>> threeRailDecrypt(threeRailEncrypt('Foo bar baz and all'))
'Foo bar baz and all'

没有zip_longest的最后一部分将是：

# recombine the triplets
plaintext = []
for i in range(len(parts[0])):
    for part in parts:
        try:
            plaintext.append(part[i])
        except IndexError:  # part too short
            pass
return ''.join(plaintext)

当导轨长度不同时，三导轨解密

1 个答案: