Vigenère Decipher

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import math
import re

def mostFrequent(List): ##Useful later on
    return max(set(List), key = List.count)

def listToString(s): ##Useful later on
    str1 = " "
    return (str1.join(s))

Alphabet = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"] ##Generally needed
Alphabet2 = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"] ##Needed to avoid problems with the .index function

ST = [] ##SecretText
Key = [] ##Key used to encode with the Vigènere Cipher
PT = [] ##PublicText
tmp = [] ##Temporary list to analyse frequency of occurence of letters

def main(file, Keylength):
    with open(file, 'r') as f:
        text = f.read().strip().split()
        len_chars = sum(len(word) for word in text) ##len_chars needed for future loops
    with open(file) as fileobj:
        for line in fileobj:
            for ch in line:
                ST.append(ch) ##SecretText into list
    for i in range(0, Keylength):
        for i in range(i, len_chars, Keylength):
            tmp.append(ST[i]) ##every i-th char decoded with the same letter in the Key
        diff = Alphabet.index(mostFrequent(tmp)) - Alphabet.index("E") ##standard frequency of occurence analysis
        Key.append(Alphabet[diff])
        del tmp[:]
    n = 0
    i = 0
    while i < len_chars:
        if n < Keylength:
            a = Alphabet.index(Key[n])
            b = Alphabet.index(ST[i])
            c = b - a
            d = c % 26
            PT.append(Alphabet2[d]) ##Reversing Vigenère
            n += 1
            i += 1
        else:
            n -= 5
    print(listToString(PT))

Note by Will Schefner
1 month ago

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ahah Nice one :D

SRIJAN Singh - 1 month ago

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