Decipher( S ) 将得到一个移位了一定数量的英文文本字符串。然后，decipher 应该尽其所能返回原始英文字符串，这将是输入 S 的一些旋转（可能为 0）。这意味着我必须尝试每种可能的解码并估计它们的英文程度。
我的方法是使用字母频率：
def letProb( c ):
""" if c is an alphabetic character,
we return its monogram probability (for english),
otherwise we return 1.0 We ignore capitalization.
Adapted from
http://www.math.cornell.edu/~mec/2003-2004/cryptography/subs/frequencies.html
"""
if c == 'e' or c == 'E': return 0.1202
if c == 't' or c == 'T': return 0.0910
if c == 'a' or c == 'A': return 0.0812
if c == 'o' or c == 'O': return 0.0768
if c == 'i' or c == 'I': return 0.0731
if c == 'n' or c == 'N': return 0.0695
if c == 's' or c == 'S': return 0.0628
if c == 'r' or c == 'R': return 0.0602
if c == 'h' or c == 'H': return 0.0592
if c == 'd' or c == 'D': return 0.0432
if c == 'l' or c == 'L': return 0.0398
if c == 'u' or c == 'U': return 0.0288
if c == 'c' or c == 'C': return 0.0271
if c == 'm' or c == 'M': return 0.0261
if c == 'f' or c == 'F': return 0.0230
if c == 'y' or c == 'Y': return 0.0211
if c == 'w' or c == 'W': return 0.0209
if c == 'g' or c == 'G': return 0.0203
if c == 'p' or c == 'P': return 0.0182
if c == 'b' or c == 'B': return 0.0149
if c == 'v' or c == 'V': return 0.0111
if c == 'k' or c == 'K': return 0.0069
if c == 'x' or c == 'X': return 0.0017
if c == 'q' or c == 'Q': return 0.0011
if c == 'j' or c == 'J': return 0.0010
if c == 'z' or c == 'Z': return 0.0007
return 1.0
我也使用这个公式：
def list_to_str( L ):
""" L must be a list of characters; then,
this returns a single string from them
"""
if len(L) == 0: return ''
return L[0] + list_to_str( L[1:] )
和这个：
def rot(c, n):
""" rot rotates c, a single character forward by n spots in the
alphabet
input: a single character
input: a non-negative integer n between 0 and 25
output: c forward by n spots in the alphabet
"""
if 'a' <= c <= 'z':
neword = ord(c) +n
if neword > ord('z'):
neword = neword - 26
elif 'A' <= c <= 'Z':
neword = ord(c) + n
if neword > ord('Z'):
neword = neword - 26
else:
neword = ord(c)
return chr(neword)
最后是这个：
def decipher( S ):
"""
"""
L = [[rot(c, n) for c in S] for n in range(26)]
LoL = [[sum(letProb(c)) for letter in encoding] for encoding in L ]
L = max(LoL)
return list_to_str(L)
前两个公式很好，但在最后一个公式中，肯定有问题：
LoL = [[sum(letProb(c)) for letter in encoding] for encoding in L ]
TypeError：“浮动”对象不可迭代