# Assistance Cracking Classical Cipher

Below is the cipher text I am trying to break and as you can see its rather short which is why I am having so much trouble.

WOYFN ZCMSH VUVTG BFUTW ABTZP FHIMF TFOSU UXFQC HKVKG MPUUQ
MHRXI OVBRZ EPJYF KKVJW GEIOV HUKEB JUNSM THIMF TFKUB ULMMQX


• Key is 6 characters long and repeats (Given by the person who encrypted it)
• There is a single character missing from the start of the ciphertext! (Also given)
• It uses a simple cipher I believe a vigenere cipher.
• Friedman IC: 1.143 (kappa-plaintext: 0.4396) which is why i think its a vigenere cipher

I have spent many hours on this but to be honest im not very good at breaking these codes.

My problem that iv run into is that once I break the cipher into its sections they are all very short and hence frequency alaysis fails.

I would really appreciate if someone could either confirm or find the crypto system being used as well as any more information about how one would find the plaintext + key or better yet find the key itself and explain how you manage it with such a short piece of ciphertext

Edit: There was some debate as to the validity of this question so here is some more information from my attempts:

Streams:

• ?ZUUZFUKUXZKEKSFU IC = 3.6833 (kappa-plaintext: 1.4167)
• WCVTPTXVUIEKIEMTL IC = 1.1471 (kappa-plaintext: 0.4412)
• OMTWFFFKQOPVOBTFM IC = 2.1029 (kappa-plaintext: 0.8088)
• YSGAHOQGMVJJVJHKM IC = 1.3382 (kappa-plaintext: 0.5147)
• FHBBISCMHBYWHUIUQ IC = 1.5294 (kappa-plaintext: 0.5882)
• NVFTMUHPRRFGUNMBX IC = 0.9559 (kappa-plaintext: 0.3676)

These numbers don't look good. they are almost all drastically off normal distribution for any language.

The main point to my question is:

• Does Vigenere Seem like a reasonable fit (given those streams are correct)?
• Is it possible to use that analysis in any way? I cant see how it can be effective with only 17 characters per stream. Most letters frequencies are 0.
• What other ciphers are possible? How could I identify that they are in use?

I am happy to provide more information I am just not sure what else is useful.

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Alas, this question is not really on topic for crypto.SE either. The FAQ is pretty clear about it. (Personally, I kind of like these kinds of questions, at least as long as they look reasonably likely to be answerable and show more research effort than just "please break this cipher for me", but the consensus seems to be that they're all off topic here.) –  Ilmari Karonen May 13 '12 at 22:48
I disagree with the fact that this isn't the place to ask. Reading the FAQ it says that its not ok to "Can I challenge people to decode something?" but it does also say that you can ask homework questions. I believe thats what im asking. I am not really after the solution more how would one approach this question given the limited ciphertext and that I have tried the attacks I know of. –  Nick May 14 '12 at 3:15
As I wrote that FAQ, I'll weigh in with what I think - the "challenge people to decode something" was meant to deal with "please decode XYZ bet you can't !11!!1"-type questions, of which we've had a few. It's not meant to discourage questions which are a decent attempt at cryptanalysis, although I think they do need to be reasonably answerable. As is, I think we could do with as much information as you can give - perhaps the results of your frequency analysis, for example? However, I don't see the need to fire a mod close vote at it, although the community or other mods may disagree :) –  Ninefingers May 16 '12 at 9:45
Also, if you feel the FAQ can be improved, do weigh in on meta :) –  Ninefingers May 16 '12 at 9:45

## migrated from security.stackexchange.comMay 13 '12 at 18:07

This question came from our site for Information security professionals.

I don't know the solution, but since you say you're only asking for hints, here's a few that occurred to me:

• If this is a Vigenère cipher, the missing character at the beginning should not matter (much): if you encrypt a message with the key FOOBAR and drop the first letter of the output, you can decrypt the resulting ciphertext with the key OOBARF.

• As ewanm89 notes, the obvious way to attack a Vigenère cipher, after you've determined the key length is $n$, is to write it with $n$ letters per row and attack each column as if it were a Caesar cipher. There are automated tools for doing that, such as pygenere; they may not always give the correct answer for very short messages, but yours looks long enough that they should at least have a decent chance.

• The most notable feature I see in your ciphertext is that the string HIMFTF occurs twice, at offsets 26 and 86. That does strongly suggest that the key length may be a divisor of 60; alas, 60 has lots of small divisors (including 6, which you believe the key length to be), so that doesn't really give that much information. It does at least rule out 7, 8 and 9 as likely key lengths.

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Hey thank you for the advice, I had noticed the HIMFTF myself. but the reason why I know the length is 6 is because the person that encrypted the message (The tutor for my class) told me that it was 6 long and that it repeats. –  Nick May 15 '12 at 0:16
PS I have now tried Pygenere again and it does not yield the message. Keep in mind that I am only guessing that this is Vigenere. Is there any way to work out if that is correct? –  Nick May 15 '12 at 0:17

The easiest way in this case to work out if it is vigenere is to brute force it.

You know the key is only 6 characters. Take a vigenere decryption function and a dictionary file of 6 character words. Decrypt using the first word, then compute a histogram of the resulting plaintext. Compare with what you would expect to see given the distribution of characters in english language text. If the two are close, visually inspect. If they are not close or visual inspection fails, continue.

Look around on google for histogram comparison ideas. A simple sum of differences should work then set a threshold for what to display.

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Cryptool is a nice crypto learning tool that supports classical cipher cryptanalysis such as for Vigenère ciphers. There is also an online version, but I'm not sure whether this only works for the German page or also for the English page. –  DrLecter Dec 17 '13 at 15:12