python - Rail Fence Cipher-寻找更好的解决方案

Question

我用 Python 编写了 Rail Fence Cipher。我想知道是否有更好的解决方案。

对于那些不知道rail fence cipher是什么的人来说，它基本上是一种以螺旋方式创建线性模式的方式编写纯文本的方法。示例 - 当“FOOBARBAZ”使用键 3 进行围栏时。

F . . . A . . . Z . . .
. O . B . R . A . Q . X
. . O . . . B . . . U .

逐行读取上面的螺旋，密文变成“FAZOBRAQXOBU”。阅读更多信息 - Rail fence - Wikipedia .

def cipher(s, key, graph=False) :
    down=True
    raw_out=[]
    out=''
    i=0
    for x in range(key) :
        raw_out.append({})
    for pos in range(len(s)) :
        raw_out[i][pos]=s[pos]
        if i==key-1 :
            down=False
        if i==0 :
            down=True
        if down :
            i=i+1
        else :
            i=i-1
    for p in raw_out :
        for q in p :
            out+=p[q]
    if graph :
        return raw_out
    return out

def decipher(s, key) :
    map_list=cipher(s, key, True) #CREATING JUST FOR MAPPING - WHICHth CHARACTER OF THE STRING - IS WHICHth CHARACTER OF THE CIPHER
    new={}
    out=''
    s_counter=0
    for x in map_list :
        for y in x :
            new[y]=s[s_counter]
            s_counter+=1
    for p in new :
        out+=new[p]
    return map_list

我想知道是否有更好的方法来执行此操作，因为我的程序非常昂贵，它使用了几个词典。

欢迎任何语言的代码。

Answer 1

只是为了它...

def fence(lst, numrails):
    fence = [[None] * len(lst) for n in range(numrails)]
    rails = range(numrails - 1) + range(numrails - 1, 0, -1)
    for n, x in enumerate(lst):
        fence[rails[n % len(rails)]][n] = x

    if 0: # debug
        for rail in fence:
            print ''.join('.' if c is None else str(c) for c in rail)

    return [c for rail in fence for c in rail if c is not None]

def encode(text, n):
    return ''.join(fence(text, n))

def decode(text, n):
    rng = range(len(text))
    pos = fence(rng, n)
    return ''.join(text[pos.index(n)] for n in rng)

z = encode('ATTACK.AT.DAWN', 3)    
print z # ACTWTAKA.ANT.D
y = decode(z, 3)
print y # ATTACK.AT.DAWN