Write a function deciphers that takes as input an arbitrary

Write a function decipher(s) that takes as input an arbitrary string s that has already been enciphered by having its characters “rotated” by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s.

This is what I got for decipher_function, however it doesn\'t work, could you please help me with decipher_function:

def rot(c, n):
\"\"\" rotates a single character c forward by n spots in the alphabet. \"\"\"
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

if \'a\'<=c<=\'z\':

new_ord=ord(c)+n
if new_ord>ord(\'z\'):
new_ord=new_ord-26
elif \'A\'<=c<=\'Z\':
new_ord=ord(C)+n
if new_ord>ord(\'Z\'):
new_ord=new_ord-26
else:
new_ord=ord(c)
return chr(new_ord)

def encipher(s,n):
\"\"\"takes as inputs an arbitrary string s and a non-negative integer n between 0 and 25, and that returns a new string in which the letters in s have been “rotated” by n characters forward in the alphabet, wrapping around as needed. \"\"\"
if s==\'\':
return\'\'
else:
rest=encipher(s[1:],n)
if len(s[0])==1:
return rot(s[0],n)+rest
else:
return 0+rest

def letter_prob(c):
\"\"\" if c is the space character (\' \') or an alphabetic character,
returns c\'s monogram probability (for English);
returns 1.0 for any other character.
adapted from:
http://www.cs.chalmers.se/Cs/Grundutb/Kurser/krypto/en_stat.html
\"\"\"
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

if c == \' \': return 0.1904
if c == \'e\' or c == \'E\': return 0.1017
if c == \'t\' or c == \'T\': return 0.0737
if c == \'a\' or c == \'A\': return 0.0661
if c == \'o\' or c == \'O\': return 0.0610
if c == \'i\' or c == \'I\': return 0.0562
if c == \'n\' or c == \'N\': return 0.0557
if c == \'h\' or c == \'H\': return 0.0542
if c == \'s\' or c == \'S\': return 0.0508
if c == \'r\' or c == \'R\': return 0.0458
if c == \'d\' or c == \'D\': return 0.0369
if c == \'l\' or c == \'L\': return 0.0325
if c == \'u\' or c == \'U\': return 0.0228
if c == \'m\' or c == \'M\': return 0.0205
if c == \'c\' or c == \'C\': return 0.0192
if c == \'w\' or c == \'W\': return 0.0190
if c == \'f\' or c == \'F\': return 0.0175
if c == \'y\' or c == \'Y\': return 0.0165
if c == \'g\' or c == \'G\': return 0.0161
if c == \'p\' or c == \'P\': return 0.0131
if c == \'b\' or c == \'B\': return 0.0115
if c == \'v\' or c == \'V\': return 0.0088
if c == \'k\' or c == \'K\': return 0.0066
if c == \'x\' or c == \'X\': return 0.0014
if c == \'j\' or c == \'J\': return 0.0008
if c == \'q\' or c == \'Q\': return 0.0008
if c == \'z\' or c == \'Z\': return 0.0005
return 1.0

def decipher(s):
\"\"\"takes as input an arbitrary string s that has already been enciphered by having its characters “rotated” by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s\"\"\"
options = [[(rot(c,n)) for c in s] for n in range(26)]
scored_options=sum(letter_prob(options))
probability=[scored_options, options, n]
return max(scored_options)
  

Solution

// In C++ Language:

string Decipher(string s, int k)
{
string str=\"\";
char c;
int i,j;
for(i=0;i<s.size();i++)
{
if(s[i]>=\'a\' && s[i]<=\'z\')
{
j=((s[i]-\'a\')-k+26)%26;
c=j+97;
str+=c;
}
else if(s[i]>=\'A\' && s[i]<=\'Z\')
{
j=((s[i]-\'A\')-k+26)%26;
c=j+65;
str+=c;
}
}
return str;
}