I wrote an attack based on frequency analysis to break the substitution cipher. The code is below by Python.


import random
import operator
import sys

cipher = """lrvmnir bpr sumvbwvr jx bpr lmiwv yjeryrkbi jx qmbm wi
bpr xjvni mkd ymibrut jx irhx wi bpr riirkvr jx
ymbinlmtmipw utn qmumbr dj w ipmhh but bj rhnvwdmbr bpr
yjeryrkbi jx bpr qmbm mvvjudwko bj yt wkbrusurbmbwjk
lmird jk xjubt trmui jx ibndt
  wb wi kjb mk rmit bmiq bj rashmwk rmvp yjeryrkb mkd wbi
iwokwxwvmkvr mkd ijyr ynib urymwk nkrashmwkrd bj ower m
vjyshrbr rashmkmbwjk jkr cjnhd pmer bj lr fnmhwxwrd mkd
wkiswurd bj invp mk rabrkb bpmb pr vjnhd urmvp bpr ibmbr
jx rkhwopbrkrd ywkd vmsmlhr jx urvjokwgwko ijnkdhrii
ijnkd mkd ipmsrhrii ipmsr w dj kjb drry ytirhx bpr xwkmh
mnbpjuwbt lnb yt rasruwrkvr cwbp qmbm pmi hrxb kj djnlb
bpmb bpr xjhhjcwko wi bpr sujsru msshwvmbwjk mkd
wkbrusurbmbwjk w jxxru yt bprjuwri wk bpr pjsr bpmb bpr
riirkvr jx jqwkmcmk qmumbr cwhh urymwk wkbmvb"""

class Attack:
    def __init__(self):
        self.alphabet = "abcdefghijklmnopqrstuvwxyz"
        self.freq = {}
        self.mappings = {}
        self.key = {}
        self.freq_eng = {'a': 0.0817, 'b': 0.0150, 'c': 0.0278, 'd': 0.0425, 'e': 0.1270, 'f': 0.0223,
               'g': 0.0202, 'h': 0.0609, 'i': 0.0697, 'j': 0.0015, 'k': 0.0077, 'l': 0.0403,
               'm': 0.0241, 'n': 0.0675, 'o': 0.0751, 'p': 0.0193, 'q': 0.0010, 'r': 0.0599,
               's': 0.0633, 't': 0.0906, 'u': 0.0276, 'v': 0.0098, 'w': 0.0236, 'x': 0.0015,
               'y': 0.0197, 'z': 0.0007}
        self.plain_chars_left = "abcdefghijklmnopqrstuvwxyz"
        self.cipher_chars_left = "abcdefghijklmnopqrstuvwxyz"
    def calculate_freq(self,cipher):
        letter_count = 0;
        for letters in self.alphabet:
            self.freq[letters] = 0

        for c in cipher:
            if c in self.alphabet:
                self.freq[c] +=1
                letter_count +=1

        for letters in self.alphabet:
            self.freq[letters] = round(self.freq[letters] / letter_count,4)

    def print_freq(self):
        for letters in self.freq:
    def calculate_matches(self):
        for cipher in self.alphabet:
            map = {} #target letter
            for plain_char in self.alphabet:
                map[plain_char] = round(abs(self.freq[cipher]-self.freq_eng[plain_char]),4)
            self.mappings[cipher] = sorted(map.items(),key=operator.itemgetter(1))
    def guess_key(self):
        for c in self.cipher_chars_left:
            for plain_char, diff in self.mappings[c]:
                if plain_char in self.plain_chars_left:
                    self.key[c] = plain_char
                    self.plain_chars_left = self.plain_chars_left.replace(plain_char,'')
        return self.key
    def set_key_mapping(self,cipher_char,plain_char):
        if cipher_char not in self.cipher_chars_left or plain_char not in self.plain_chars_left:
            print("ERROR: No Mapping Exists!",cipher_char,'__',plain_char)
        self.key[cipher_char] = plain_char
        self.cipher_chars_left = self.cipher_chars_left.replace(cipher_char,'')
        self.plain_chars_left = self.plain_chars_left.replace(plain_char,'')

def decrypt(key,cipher):
        message = ""
        for c in cipher:
            if c in key:
                message += key[c]
                message += c
        return message

attack = Attack()

#mapping improvement


key = attack.guess_key()
message = decrypt(key,cipher)

cipher_lines = cipher.splitlines()
plain_lines = message.splitlines()
for i in range(len(cipher_lines)):
    print('M: ',plain_lines[i])
    #print('C: ',cipher_lines[i])

... Mission Done

This case it works really well, but with a different substitution encryption, for example, one such that it doesn't prefer to encrypt based on frequency analysis in English. Thus, wut I wanna ask, is that, if the frequency analysis fails in some of the cases, wut's one kind of idea to improve the attack(required to use frequency analysis as the main attack scheme)? Appreciate if someone can help!


2 Answers 2


This is the problem. In most of cryptography textbook (ciphertext above is from Paar's book) authors show frequency analysis as main tool. In reality it is not enough. This book Solomon Kullback STATISTICAL METHODS IN CRYPTANALYSIS is one of the best resourses. (Sinkov's book is other). It included vowel/consonant recognition and other methods.


problem statements


Large corpora are available to us, each written in one of K different natural languages, which use a Roman-like alphabet (perhaps Greek, Cyrillic, ...).

Plaintext is produced in one of those languages, with roughly typical word frequencies (so "the" is ranked first in English).

When a substitution cipher is applied, non-letter glyphs (punctuation) are dropped, and accented letters are mapped to one of 26 letters ("déjà" --> "DEJA").


Problem 1: Given a cipher text, recover the plaintext and its language.

Problem 2: Given a cipher text with no SPACE characters, do the same.


common word lengths

The OP is an instance of Problem 1, and it features many occurrences of "w", "jx" and "bpr". For a blind language guess of "English", this immediately suggests trial decryption with (w -> a, b -> t, p -> h, r -> e). When this doesn't work out we can fall back to (w -> i), and so on. At this point we're essentially playing Hangman, making wildcard matches against a dictionary.


Given enough input text, we can expand our frequency table to digrams and even trigrams. This works even for Problem 2, where word boundaries are not available.

Each of the K languages has its own etaoin shrdlu Zipfian distribution. We can maybe say that most vowels will appear within the top ten, but beyond that the unigram frequencies don't really help us recover the language identifier.

In contrast, rank ordering the cipher digrams is very helpful for identifying the language, assuming we're presented with a large enough sample. Iterate through each language, and then through each digram by descending frequency, performing trial substitutions. To evaluate we need only consult the digram ranking, and not the ciphertext. Is the substitution consistent with other digram frequencies? Or would it induce too much loss, perturbing the ranking by too much? Accept or reject the substitution, trying to minimize total loss.

This parallelizes nicely, producing an evolving rank ordering of which are the most likely handful of source languages, so early in the search we focus CPU cycles on the most promising ones. And within a language guess, we're back to playing Hangman.


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