hribeirosantana · May 2, 2019 07:04 · yashgupta1262 · Jul 11, 2019 · tanzz1337 · Sep 28, 2019
diff --git a/rxzryiyxyion.py b/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()