December 20th, 2009, 04:51 PM
Simple XOR theory
Hey guys I have always wondered this but no one's been able to answer me. I know how to crack a simple XOR (not a one-time pad, but one where the key length is smaller than the plain text).
According to Applied Cryptography Second Edition you are supposed to...
How does this actually work? What's the mathematical theory behind it?
December 21st, 2009, 12:27 AM
Yes it works, XOR is the simplest cipher.
Thing XOR key => ciphertext
ciphertext XOR key => plain text again.
The technique simply helps you find the key. If the key is shorter than the clear text, then it repeats. Find that, and you are done.
December 22nd, 2009, 01:59 PM
Take two random characters from a text. Compare them. What is the probability that they will be the same character? One divided by the number of characters in the alphabet.
Originally Posted by Number1Brat
If, for example, your text was WWI-era English telegraph cipher interceptions, you'd have an alphabet of 26 characters (upper-case A-Z with no spaces), so with random text you'd expect a match - a coincidence - 1/26, or 3.85% of the time.
Now, suppose your text is standard English in the same alphabet. What would be the odds? Much higher. The precise value, like the value for random text, depends upon the alphabet, and it also depends on the language. Ordinary English, written as a telegraphic cipher (upper-case A-Z with no spaces), yields a coincidence 6.65% of the time.
These values are referred to as the Indices of Coincidence of random text and of plain text.
Now the interesting thing about monoalphabetic ciphers is that they don't change the index of coincidence (IC). Substituting 'X' for 'A' throughout a text doesn't change the likelihood that the two characters will be the same. So a monoalphabetic cipher will have the same IC as plaintext.
But polyalphabetic ciphers are very different. Each of the individual alphabets will have the same IC as plaintext, but because the different alphabets are all muddled together, their combined IC will be much closer to that of random text than of plain text. In other words, the IC of the cipher text will be close to 0.0385.
Now, what happens when you subtract the text from itself, shifted by one or more places. If the amount of the shift is unrelated to the length of the keyword, each letter is being subtracted from a letter in a different alphabet, and the result is still pretty much random, so you should see 0 about 3.85% of the time. But when the shift is a multiple of the length of the keyword, each letter is being subtracted from a letter that is in the same alphabet it is. So you'd expect to see 0 about as often as two random characters from plaintext would match. That is, 6.65% of the time.
Think about it. You have two ciphertext letters, each of which is the sum of a key letter and a plaintext letter:
C1 = P1 + K1
C2 = P2 + K2
When you subtract the two ciphertext letters, you get:
C2-C1 = P2-P1 + K2-K1
But when the two letters are separated by a multiple of the keylength, K1 and K2 are the same, and you get:
C2-C1 = P2-P1 + K1-K1
C2-C1 = P2-P1
A value that is independent of the key.
When you're done with the process, you have determined the length of the key. With that, you can split the message into its separate alphabets, and attack each as an ordinary monoalphabetic substitution. Or, you might recognize that the result of your subtract is the plaintext subtracted from itself, shifted by the length of the key. In other words, it's an autokey cipher, which has known methods of attack.
Comments on this post
September 6th, 2011, 05:40 PM
please . can I uses these ways ..if I have plaintext in hex xor key in hexa and the result naturally in hex..in this case can I separate the key from plaintext and return the original text