使用排列作为键解扰消息字符串

Question

帮助我正在尝试使用排列作为密钥对文件进行解扰，我知道如何对其进行加扰，但我需要创建一个函数来将其解扰回使用相同排列密钥的内容，这是代码：

def stringtoarray(S):
    A = []
    i = 0
    while i<len(S):
        A.append(S[i])
        i += 1
    return A

def arraytostring(A):
    S = ""
    for c in A:  # "for each element/value c in array A"
        S = S+c
    return S
#arraytostring
def scramble(S, P):  #Scramble string S with permutation P
    pn = len(P)  # length of permutation array
    E = (len(S) + pn - (len(S)%pn)) * [' ']  # array of chars, padded 
    i = 0
    while i< len(S):
        seg = i/pn   # segment number
        j = i % pn   # segment offset
        E[ seg*pn + P[j] ] = S[i]
        i += 1
    # while
    return arraytostring(E)
# scramble
print scramble("0123456789abcdefghij",[9, 14, 11, 19, 16, 18, 12, 6,
               7, 15, 0, 5, 17, 4, 3, 10, 2, 1, 8, 13])

# prints ahgedb78i0f26j194c53

我想再次将其设置回字符串“0123456789abcdefghij”

Answer 1

从某种意义上说，您已经编写了解扰器。你所要做的就是喂scramblerinverse permutation。下面，逆排列由argsort(P) 给出，其中P 是排列。

---

def scramble(S, P):  #Scramble string S with permutation P
    pn = len(P)  # length of permutation array
    E = (len(S) + pn - (len(S)%pn)) * [' ']  # array of chars, padded
    for i in range(len(S)):
        seg,j = divmod(i,pn)
        E[ seg*pn + P[j] ] = S[i]
    return ''.join(E).rstrip()  

def argsort(seq):
    # http://stackoverflow.com/questions/3382352/3382369#3382369
    '''
    >>> seq=[1,3,0,4,2]
    >>> index=argsort(seq)
    [2, 0, 4, 1, 3]

    Given seq and the index, you can construct the sorted seq:
    >>> sorted_seq=[seq[x] for x in index]
    >>> assert sorted_seq == sorted(seq)

    Given the sorted seq and the index, you can reconstruct seq:
    >>> assert [sorted_seq[x] for x in argsort(index)] == seq
    '''
    return sorted(range(len(seq)), key=seq.__getitem__)


P=[9, 14, 11, 19, 16, 18, 12, 6, 7, 15, 0, 5, 17, 4, 3, 10, 2, 1, 8, 13]
text="0123456789abcdefghij"
print scramble(text,P)
# ahgedb78i0f26j194c53                    
print(scramble(scramble(text,P),argsort(P)))
# 0123456789abcdefghij                        

---

顺便说一下，不要为E预先分配足够的空间

E = (len(S) + pn - (len(S)%pn)) * [' ']

您可以使用argsort按顺序生成E中的项目：

def scramble(S, P):  
    pn = len(P) 
    E = []
    idx = argsort(P)
    for i in range(len(S)):
        seg,j = divmod(i,pn)
        E.append(S[idx[j]+seg*pn])
    return ''.join(E)

Answer 2

简单。计算排列的倒数并应用它。有一个众所周知的算法。你可以发现它被引用为 Donald Knuth 的“计算机编程艺术”中的算法 J。

这里似乎有源代码：http://binetacm.wikidot.com/algo:perminv

void inversePermutation(int perm[], int n, int inv[]) {
   for (int i=0; i<n; i++)
      inv[perm[i]]=i;
}

Answer 3

首先在您的代码中添加一些 cmets。

以下是完全没有必要的：

def stringtoarray(S):
    A = []
    i = 0
    while i<len(S):
        A.append(S[i])
        i += 1
    return A

只要做list(your_string)就可以实现

另外，如果你想迭代一些东西，你可以使用for .. in .. 例如。

for char in some_string:
    # do something with character

您的arraytostring 函数也是不必要的。将列表连接成字符串的正确方法是使用string.join()^[docs]。例如，''.join(['a', 'b', 'c']) 会给你'abc'。

---

这就是我将如何执行加扰/解扰功能：

def scramble(s, p):
    chars = list(s)
    scrambled = ['']*len(chars)
    for char, i in zip(chars, p):
        scrambled[i] = char
    return ''.join(scrambled)

def unscramble(s, p):
    reverse_p = sorted(range(len(p)), key=p.__getitem__)
    return scramble(s, reverse_p)

s = "abcdefg"

p_key = [6, 1, 3, 0, 5, 2, 4]

print "original:", s

scrambled = scramble(s, p_key)

print "scrambled:", scrambled

print "unscrambled:", unscramble(scrambled, p_key)

结果：

原文：0123456789abcdefghij 加扰：ahgedb78i0f26j194c53 解读：0123456789abcdefghij