hribeirosantana/rxzryiyxyion.py

## rxzryiyxyion.py
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Apr 29 04:25:03 2019

@author: hugo
"""

"""This is a FAILED attempt to solve evilzone's challenge "encryption 2"

"Figure out what this says:
rxzryiyxyion wiqhasr wen za zsoman feislt aerilt, er lonk er yha wiqhas
yauy ir lonk anoxkh, end wonyeinr anoxkh woggonlt xrad whesewyasr."

The attempt assumes various things about the ciphered text (e.g.
the substitution is the same for equal letters, the text is not
reversed, it is in english, etc). The idea was (1) to pick a slice of
the ciphered text (I picked "er lonk er yha") in which it contained
a sequence of small words (two, three, and four letters words),
(2) research for what are the most frequent words in english of
that size, (3) build all possible permutations using those words,
(4) compare each one of those permutations against some
big text file (like a portion of Wikipedia's backup or the bible),
and then, (5) hopefully seeing what were the most common sentences formed
with of those words, (6) simply manually substitue the letters in
the ciphered text hoping that it would give us a clue about the rest.

Usage: simply run it in the same directory of your file.txt.

"""

FILE = "file.txt" # change to your big text file
SENTENCE_SIZE = 14 # er lonk er yha
ONE_BYTE = 1
ZERO = 0

two_letter_words = ['of', 'to', 'in', 'it', 'is', 'be', 'as', 'at', 'so', 'we']
three_letter_words = ['the', 'and', 'for', 'are', 'but', 'not', 'you', 'all', 'any', 'can']
four_letter_words =  ['that', 'with', 'have', 'this', 'will', 'your', 'from', 'they', 'know', 'want']

def build():
    sentences_set = set()
    for x in two_letter_words:
        for y in three_letter_words:
            for z in four_letter_words:
                sentence = x + ' ' + z + ' ' + x + ' ' + y
                sentences_set.add(sentence)
    return sentences_set

def evaluate(sentences_set):
    sentences_dict = dict()
    counter = 1
    for sentence in sentences_set:
        print('Evaluating sentence ' + str(counter))
        result = count_ocurrences(sentence)
        sentences_dict[sentence] = result
        counter += 1
    return sentences_dict

def count_ocurrences(sentence):
    ocurrences = ZERO
    offset = ZERO
    with open(FILE, 'r') as file:
        while True:
            file.seek(offset)
            text = file.read(SENTENCE_SIZE)
            if not text:
                break
            if sentence == text:
                ocurrences += 1
            offset += ONE_BYTE
    return ocurrences

def print_answer():
    sentences_set = build()
    sentences_dict = evaluate(sentences_set)
    for element in sentences_dict.items():
        print(element)

if __name__ == "__main__":
    print_answer()
	#!/usr/bin/env python3
	# -- coding: utf-8 --
	"""
	Created on Mon Apr 29 04:25:03 2019

	@author: hugo
	"""

	"""This is a FAILED attempt to solve evilzone's challenge "encryption 2"

	"Figure out what this says:
	rxzryiyxyion wiqhasr wen za zsoman feislt aerilt, er lonk er yha wiqhas
	yauy ir lonk anoxkh, end wonyeinr anoxkh woggonlt xrad whesewyasr."

	The attempt assumes various things about the ciphered text (e.g.
	the substitution is the same for equal letters, the text is not
	reversed, it is in english, etc). The idea was (1) to pick a slice of
	the ciphered text (I picked "er lonk er yha") in which it contained
	a sequence of small words (two, three, and four letters words),
	(2) research for what are the most frequent words in english of
	that size, (3) build all possible permutations using those words,
	(4) compare each one of those permutations against some
	big text file (like a portion of Wikipedia's backup or the bible),
	and then, (5) hopefully seeing what were the most common sentences formed
	with of those words, (6) simply manually substitue the letters in
	the ciphered text hoping that it would give us a clue about the rest.

	Usage: simply run it in the same directory of your file.txt.

	"""

	FILE = "file.txt" # change to your big text file
	SENTENCE_SIZE = 14 # er lonk er yha
	ONE_BYTE = 1
	ZERO = 0

	two_letter_words = ['of', 'to', 'in', 'it', 'is', 'be', 'as', 'at', 'so', 'we']
	three_letter_words = ['the', 'and', 'for', 'are', 'but', 'not', 'you', 'all', 'any', 'can']
	four_letter_words = ['that', 'with', 'have', 'this', 'will', 'your', 'from', 'they', 'know', 'want']

	def build():
	sentences_set = set()
	for x in two_letter_words:
	for y in three_letter_words:
	for z in four_letter_words:
	sentence = x + ' ' + z + ' ' + x + ' ' + y
	sentences_set.add(sentence)
	return sentences_set

	def evaluate(sentences_set):
	sentences_dict = dict()
	counter = 1
	for sentence in sentences_set:
	print('Evaluating sentence ' + str(counter))
	result = count_ocurrences(sentence)
	sentences_dict[sentence] = result
	counter += 1
	return sentences_dict

	def count_ocurrences(sentence):
	ocurrences = ZERO
	offset = ZERO
	with open(FILE, 'r') as file:
	while True:
	file.seek(offset)
	text = file.read(SENTENCE_SIZE)
	if not text:
	break
	if sentence == text:
	ocurrences += 1
	offset += ONE_BYTE
	return ocurrences

	def print_answer():
	sentences_set = build()
	sentences_dict = evaluate(sentences_set)
	for element in sentences_dict.items():
	print(element)

	if __name__ == "__main__":
	print_answer()