加密柱状转置密码 [英] Encrypting a columnar transposition cipher

问题描述

我试图找出如何加密Python中的柱状转置密码，给出明文大写字母和任何长度的数字键。例如，如果键为3124，字符串为IHAVETWOCATS，则将按如下方式组织字符串：

I'm trying to figure out how to encrypt a columnar transposition cipher in Python given a plaintext uppercase string and a number key of any length. For example, if the key is 3124 and the string is 'IHAVETWOCATS', it would organize the string like so:

3124
IHAV
ETWO
CATS

然后返回列中的字符1，然后列2等，直到最后返回加密的字符串'HTAAWTIECVOS'。到目前为止，我知道我需要使用一个累加器，我一直在使用一个字典，但是我完全陷入困境。这些是我尝试过的一些功能：

and then return the characters in column 1 first, then column 2, etc, until finally returning the encrypted string 'HTAAWTIECVOS'. So far I know that I'll need to use an accumulator, and I've been toying with the idea of using a dictionary, but I'm just completely stuck. These are some of the functions I've tried:

def columnar(plaintext,key):
    cipher=''
    acc=0
    for i in range(len(key)):
        while acc<(len(plaintext)/len(key)):
            cipher=cipher+plaintext[i+acc*5]
            acc=acc+1
    return(cipher)

^只返回几个字母，而不是一个适当长度的字符串。

^This only returns a few letters, not a string of appropriate length.

def columnar（明文，键）
值= {}
seqlist = []
nextvalue = 1
indices = rand（len（key））
为明文中的字母：
为i在索引：
if letter == key [i]：
values [i] = nextvalue
nextvalue = nextvalue + 1
for i in indexes：
seqlist.append值[i]）
返回seqlist

^上述函数返回一个KeyError：0错误。
非常感谢您的帮助！

^The above function returns a KeyError: 0 error. Thank you very much for any help!

推荐答案

def split_len(seq, length):
    return [seq[i:i + length] for i in range(0, len(seq), length)]

def encode(key, plaintext):

    order = {
        int(val): num for num, val in enumerate(key)
    }

    ciphertext = ''
    for index in sorted(order.keys()):
        for part in split_len(plaintext, len(key)):
            try:
                ciphertext += part[order[index]]
            except IndexError:
                continue

    return ciphertext

print(encode('3214', 'IHAVETWOCATS'))
#>>> HTAAWTIECVOS

split_len 是Ian Bicking

split_len is by Ian Bicking

所以我用 split_len 将代码分成块，然后使用字典理解来获取索引的顺序，该订单中的字母。

So i split the code into chunks with split_len then use dictionary comprehension to just get correct order of indexes and finally i concatanate the letters in that order.

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