I love challenges of all kinds, both the algorithmic and cryptographic. There is just something inherently compelling about pitting your wits alone against an intellectual challenge, free to apply any and all means to beat it. I recently took part in a crypto challenge where we were meant to break a text encoded with the running key cipher, which I hadn’t encountered before, and I came up with a rather satisfying solution for I thought might be good to share here.

For those of you who aren’t familiar with the running key cipher, it’s very simple. Say you have some text you would love to be kept secret:

IENJOYEATINGWHILESUSPENDEDUPSIDEDOWN

But still feel the need to communicate it to someone, with whom you have agreed some (usually well known) key beforehand, say:

GOBOILYOURBOTTOMSYOUSONSOFSILLYPERSONS

We just “add” the two texts together, with the letters being used as numbers in the obvious way: A = 0, B = 1, …, Z = 25. This means that e.g. C + B = 2 + 1 = 3 = D. Furthermore, we do this modulo 26, so that e.g. Z + B = A. Additionally, since in our case our key is a bit longer than the secret message, we will just not use the last couple of letters of the key. This means that our encoded message is:

OSOXWJCONZOUPAWXWQIMHSAVSIMXDTBTHFOB

The receiver can just subtract their known key to retrieve the original text. Wow, great! Since we never never reuse the key at any point in that message, and assuming we never use they key again, it must constitute a one time pad! This means that we must have perfect security, right? Ah, grasshopper, if it only it were so simple! Actually, this is not very secure at all because both our key and our message exhibit distinctly non-random distributions of letters. A simple example of this can be seen if you imagine that we subtract just the letter E (the most commonly used in English) from every letter in the intercepted ciphertext. We would expect to get considerably more than 1/26 of the resulting letters being either from the key or plaintext. If we do this with our example, we obtain:

KOKTSFYKJVKQLWSTSMEIDOWROEITZPXPDBKX
K    K       K      K  K      KP

I’ve labelled each resulting letter with a K or a P depending on whether it came from the key text or the plain text. As you can see, most of the letters are from the key because it contains very few E characters.

This is all very well, but it’s clearly a fairly crude attack. If we want to have a chance of actually breaking the cipher we need some more heavyweight methods. As far as I know, there are a choice of three major techniques:

  • Known plaintext attacks
  • Crude N-gram analysis
  • Hidden Markov model N-gram analysis

The last one is what I actually used to break the challenge, and the one that I think is just plain coolest, but we need to build up to it as it’s pretty scary :-). So, let’s take a look at the other techniques first:

Known Plaintext Attack
This is definitely the easiest attack to employ if you know something about the message being sent. For instance, if you, as the evil attacker, happen to know that the sender is a passionate gourmand, you might guess that EATING will appear somewhere in the key or plaintext. Knowing this, you can exhaustively subtract EATING from the ciphertext at every offset. If we do this for our example text, we obtain:

Text offset Text deciphered with key
0 KSVPJD
1 OOEOWW
2 KXDBPI
3 TWQUBH
4 SJJGAT
5 FCVFMI
6 YOURBO
7 KNGGHJ
8 JZVMCU
9 VOBHNQ
10 KUWSJR
11 QPHOKQ
12 LADPJK
13 WWEODC
14 SXDIVG
15 TWXAZB
16 SQPEUM
17 MITZFU
18 EMOKNP
19 IHZSIM
20 DSHNFC
21 OACKVG
22 WVZAZR
23 RSPEKX
24 OITPQN
25 EMEVGV
26 IXKLON
27 TDATGB
28 ZTILUZ
29 PBAZSI
30 XTOXBV

Most of those look like gibberish, but one stands out as eminently readable English text: YOURBO. This is, of course, part of the key text. The diligent cryptanalyst will then typically be able to use further guesses (perhaps inspired by the discovered piece of text) to reveal more of the message and key. For instance, he might try every word beginning with BO to try and extend the known plaintext rightwards (though this is a rather difficult proposition since there are at least 368 possibilities according to my dictionary). A small aside here about my methodology: I found the optimal experience was got by combining a suitable dictionary (I use 5desk) with grep in a Cygwin Bash shell. You can then make queries such as the one below, which counts the number of words in the dictionary beginning with “bo”.

$ grep ‘^bo’ 5desk.txt | wc -l
368

Note that it is not strictly possible to tell whether the English words you recover are part of the key or plain text. This is because of the symmetry of the modular addition we got to perform the ciphering step. However, it is usually possible to work it out by your knowledge of the messages probable contents. Once you know which text stream is part of the key, a Google search will usually allow you to obtain the entire thing from your fragmentary knowledge, assuming that the key is a public text (as is typical: Bible extracts are popular). This will let you break the code!

However, what if you don’t know anything about the message, or, like me, you’re just a bit rubbish at making good guesses as to how to extend your known fragment of text? We probably need to employ something a little more sophisticated, like one of the N-gram analyses I mentioned earlier. However, as this blog entry is already getting overlong I shall leave exposition of these rather more clever techniques for a future entry: watch this space, especially those of you whom I have just bored senseless by explaining the absolute basics.

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