XOR Encryption uses the XOR operator ( Exclusive Or, symbol: ⊕) with the plain text and the key as operand (that should be binary encoded).Example: The plain message is 1001 and the key is 10Take the first bit ( 0 or 1) of the plain text and the first bit of the key and then using XOR operation to get the ciphered bit.Example: 1 ⊕ 1 = 0The operation is repeated with the second bit of the plaintext and the second bit of the key. At the end of the key, loob back to the first bit.Example: Plain message1001Key (repeated)1010Encrypted message0011. XOR is comptabile with the principle of an ciphering. Ask a new questionSource codedCode retains ownership of the source code of the script XOR Cipher online.
Which works just fine, like: string ciphertext = XOR('test', '1234'); will return the ciphertext and when decrypting:string plaintext = XOR(ciphertext, '1234'); it will return 'test'. So, I wanted to create an algorithm that would break the xor cipher by brute forcing, so basically trying to decrypt the ciphertext with every possible key. Encryption / Decryption tool » Online Encrypter / Decrypter tool. This program let you encode/decode or encrypt/decrypt your documents in various formats such as: ASCSII, Binary, Base 64, Hex, Caesar Bruteforce, MD5, SHA-1, HTML Entities.
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People who know me would also know how much I have been looking forward to this exercise. Of is about encryption, and modern days encryption relies heavily on mathematics. I had much fun trying to crack the code of this problem. The problems reads:Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (.) = 42, and lowercase k = 107.A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes.
The user would keep the encrypted message and the encryption key in different locations, and without both “halves”, it is impossible to decrypt the message.Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.Your task has been made easy, as the encryption key consists of three lower case characters. Using (right click and ‘Save Link/Target As’), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.We could brute force through all the keys since there are only 26 3 = 17576.
However, wWe know quite a lot about the encrypted message, but lets just have a short list of what we actually know about the encrypted text. The text is in English. The key has three digits. All digits are lower case English letters.
We know the encryption method. Hi SuprDewdThanks for letting me know about extension methods they are hugely helpful and a way to implement something I would need over and over again. I played around with this problem and reduced the analysis code to a few lines with linq.Regarding the decryption method and it’s real life applications.
It wouldn’t work on any modern day mathematical encryption methods. It only works methods like this where you could basically perform the method by hand.
However for a few thousand years encryption was based on methods like the one used in this problem./Kristian. Hi GuyThanks for that question, which means I haven’t explained it well enough. So here is a bit of elaboration on the code.There are two sections of this code. The first section going from line 1 to 6.This section loops through the message one character at the time and counts the occurrence of each character. However, it splits it into three sections since the key length is known to be 3. Which section to increase the count in is controlled by line 3.
Line 4-5 keeps track of the most common character found so far for each key index.The splitting into multiple piles is also explained in the post as: But our key is not one letter, ours is known to be three letters. However, we can split the encoded message into three piles. One encrypted by the first character of the key. This pile will be the characters 1,4,7,10 The second pile will be every third letter starting from the second, and so on. On each of these piles we do frequency analysis as described above and then we are likely to get the key for each pile.In the second section I then generate the key. I then assume that the a space is the most commonly occurring character in the text, so for each of the most commonly occurring character I xor it with the value for spacebar (ascii value 32).
![Xor Xor](https://www.itwire.com/media/k2/items/cache/fbb11dcf0cc5c42ea0d8ee855e698780_XL.jpg)
So now I have the three character key ready to apply to the message.Does this make sense? At least I hope it does, otherwise just ask again./Kristian. This is one of those problems that python seems particularly well suited to. If we assume our most frequent character is space, then we slice the cipher text into 3 buckets of every 3rd character and xor the most frequent character in each bucket with space to generate the key. Then it’s a simple matter to produce the decrypted text.Solution runs in roughly 1.4 ms. Much faster and more concise than other approaches I’ve seen.#!
/usr/bin/env pythonimport itertoolsfrom collections import Counterdef solve(filename='cipher1.txt', keylen=3):cipher = map(int, open(filename).read.split(','))key = map(lambda n: max(Counter(ciphern::keylen).iteritems,key=lambda x: x1)0 ^ ord(' '),xrange(keylen))return sum(c^k for c, k in itertools.izip(cipher, itertools.cycle(key)))if name 'main':print solve.