You found 2 messages on a folded piece of paper.

Osmz bdajamu se l qxufkw zvfciodrf zss, wq xv wrz kje zq jni nfla ut lrfdmm. Thkic bxswibi ep jniojei tzrhil kt snafq tp of gjhwp, pxa naoz qxp bu we bhat i ukp ma copum bvcs sqkqequ.

Ywel xfkbaao yg p sdsjym lnviiynzd diq, om jh eln mzi rw dve xdxa id pjrzow. Lhyci dziaahc mv dveyhwe jjvfax qt mbifu hv ed evbef, lfs jcaj cfn pq qy jjan s yqh ac culki ltos gsauwwo.

Interested, you want to decode these 2 messages

What are the messages, and what is the key?

Hint 1:

These 2 ciphers are the same message, encoded 2 different ways using the same key.

Hint 2:

Vigenere is the first one

Hint 3:

The second one is the opposite. Decode with the key in Vigenere to encode what is known as Variant Beaufort. Encode with the key (in Vigenere) to decode Variant Beaufort.

Hint 4:

The last word is message, if that helps.

I have noted that the key is 6 alphabets long, so A PARTIAL KEY is all that's needed

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    $\begingroup$ You should probably check out meta.puzzling.stackexchange.com/questions/5005/… - you should give people a chance to solve your puzzle without hints first (plus four is a lot, and you basically gave us instructions to solve it :/) $\endgroup$ – user812786 Oct 20 '16 at 19:13
  • $\begingroup$ To turn sqkqequ into message the key would have to be gmsyekq. According to a Kasiski's Examination I performed the only repitition in the ciphertext is jni with a gap of 29; This also suggests a key length of 29, but it might be a coincidence. Here are all possible key formats that keep "message", where # represents an unknown character. The only one with a length of 29 has this format: ###################gmsyekq###. So I guess now it's just guessing what the words are? gmsyekq seems so random to me... (maybe steganography is involved?) $\endgroup$ – user14478 Oct 20 '16 at 19:17
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    $\begingroup$ Btw. this is definitely not a Vigenere cipher. Vigenre uses a password repeated over and over, which is obviously not the case here. This is rather a one-time pad. $\endgroup$ – Sleafar Oct 20 '16 at 19:28
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    $\begingroup$ @LukasRotter If you look at both ciphertexts, there are places where both of them contain the same letter at the same place. At this places the letter of the message would be the same as in both ciphertexts, and the letter of the key would be an "A". Looking at the distances of all these "A's" I can't see any possible password shorter than the text. However, it should be still solvable, because there are only 2 possible letters at each place, but it's too late for me now to do this today. $\endgroup$ – Sleafar Oct 20 '16 at 19:39
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    $\begingroup$ @Lukas Basically he's given us two ciphertexts $C_{+}=P+K\pmod{26}$ and $C_{-}=P-K\pmod{26}$; therefore the plaintext is just $P=(C_{+}+C_{-})/2\pmod{26}$ (with the complication that, because $26$ and $2$ are not coprime, the division creates two possibilities for each letter). $\endgroup$ – 2012rcampion Oct 20 '16 at 20:00

See my comment on the question for the method; the plaintext is:

This message is a rather sensitive one so do not let it get seen by anyone. Chess playersisare getting better at chess as we speak, now your job is to find a way to crack this message.

(How we were supposed to read those instructions before getting the plaintext is anyone's guess.)

The key is:


(Which is consistent ($p>0.1$) with letters drawn at random from a uniform distribution.)

  • $\begingroup$ You are correct, both in the message AND the partial key. $\endgroup$ – garr890354839 Oct 21 '16 at 12:41

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