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| Double cipher challenge | |
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| Topic Started: Aug 10 2007, 12:23 PM (693 Views) | |
| catish10 | Aug 10 2007, 12:23 PM Post #1 |
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Just registered
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Hello everyone! Please, please please can I have some help!!
 I have found a code on the internet and I am trying to break it. Here it is.FUKPGCZHZCI MVZKOC, QBH ZW DTS KR QGBW GM TCIDZTN MVXCWLGAZ QHDSK YP RMKDL RYUTSPQ KDPC NSAGDKHZO PX CYRQDCK AQ AG IBZ P GRIU VBMQ CAK PVOHLBZS WDM RWSDIFAYVX VCKL ABM RHOZGIS ! ST FRDP WQUG. PK UTH STBA INCS PK PT RPRQU UCT CL Z WEBW AYPW XBAOREM JGGUH PS RUO WNM ZRTWDMC BZWDPVRCG PA ZSI PGPV. HSPX SFVI W RFA TTYG NFICKQYKH I have been told that the final answer is in English and that there are two systems used. I have been at this almost a week, and all I've managed to do is caluculate the IC (I get a spike at 6 and 12) and I have done a Kasaki test ( repeat pattern factor looks like 6 for a key word) . Frequency count shows all 26 letters used (possibly rules out playfair). I am thinking substitution (keyed ?) followed by Vigenere. The thinking behind this is that there are three single letters which when decrypted can only be A or I but in the cipher text they have three different letters. The only way I know to do this is Vigenere. Please can someone have a look and let me know what they think. I dont want the answer - I like a challenge - but I have not worked with layered ciphers, and I REALLY want to learn as much as possible. Any help would be greatfully received. (PS this is not a competition, only a challenge!)  Thanks
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| jdege | Aug 11 2007, 06:43 PM Post #2 |
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Think about it a bit more. Consider what happens when you apply a substitution to the output of a substitution. The result is just another substitution. That is, if in the first substitution, A=>X, and in the second X=>G, then you've really done nothing more that A=>G. The same holds true no matter how many different substitutions you apply - there is always a single substitution that produces the same result. Multiple passes of Vigenere using different keys, but of the same length, is the same. Encrypt once with keyword FRED and then again with the keyword ADAM, and the result is the same as if you'd encrypted with the keyword FUEP. Now think about what happens, if you encrypt with a simple substitution, then again with Vigenere - the result is the same as if you'd encrypted with Vigenere alone. The keyword for this equivalent Vigenere is different than what you used in the original, but that doesn't matter to the cryptanalyst, who doesn't know the keyword anyway. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| catish10 | Aug 12 2007, 08:53 AM Post #3 |
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Hi Thanks, I think I understand you. If its encrypted with a sub + vig = vig with a strange keyword! Have you broken it? If I am correct in thinking this is a sub+vig to break it I just need to treat it as a normal vig. So this gives me at least three different ways of attacking it! Can you confirm I am right with a keyword length of 6? Michelle
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| Revelation | Aug 12 2007, 02:49 PM Post #4 |
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Well, you can never be sure. My IOC calculator gives the highest peak at twelve:
This means that it is most likely 6 or 12. You said that the kasaki test yielded 6, so that's probably the keylength. |
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RRRREJMEEEEEPVKLWENFNVJKEEEEEAOLKAFKLXCFZAASDJXZTTTTTTTLSIOWJXMOKLAFJNNKFNXN RAGRBAQEMHIGDJVDSEOXVIYCELFHWLELJFIENXLRATALSJFSLCYTKLASJDKMHGOVOKAJDNMNUITN RRRRLJVEEEEECLYVYHNVPFTAEEEEEMWLMEIRNGLARWJAKJDFLWNTIERJMIPQWOTZEOCXKNUBNXCN RJIRPOWEANFUSNCZVDVZNMSFEKLOEPZLDKDJWSAAAAAAAOERHJCTNCKFRIMVKSOFOMKMANREWNBN RZUDRGXEEEEENFQIDVLQNCKNEEEEEDGLLLLLLAWIOSNCDARLODMTOEJXMILDFJROTKJSDNLVCZNN | |
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| Revelation | Aug 12 2007, 03:41 PM Post #5 |
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I can tell you that it is not 12. Though, I am not able to solve it with 6 either. There probably is another encryption. Maybe a transposition. |
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RRRREJMEEEEEPVKLWENFNVJKEEEEEAOLKAFKLXCFZAASDJXZTTTTTTTLSIOWJXMOKLAFJNNKFNXN RAGRBAQEMHIGDJVDSEOXVIYCELFHWLELJFIENXLRATALSJFSLCYTKLASJDKMHGOVOKAJDNMNUITN RRRRLJVEEEEECLYVYHNVPFTAEEEEEMWLMEIRNGLARWJAKJDFLWNTIERJMIPQWOTZEOCXKNUBNXCN RJIRPOWEANFUSNCZVDVZNMSFEKLOEPZLDKDJWSAAAAAAAOERHJCTNCKFRIMVKSOFOMKMANREWNBN RZUDRGXEEEEENFQIDVLQNCKNEEEEEDGLLLLLLAWIOSNCDARLODMTOEJXMILDFJROTKJSDNLVCZNN | |
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| catish10 | Aug 12 2007, 06:09 PM Post #6 |
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Just registered
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Hi I think I can rule out substitution followed by transposition because of the frequencies Histogram Analysis of <Unnamed1>. File size 329 bytes. Descending sorted on frequency. No. Substring Frequency (in %) Frequency 1 P 7.3077 19 2 C 6.1538 16 3 R 5.7692 15 4 G 5.3846 14 5 K 5.3846 14 6 S 5.3846 14 7 Z 5.3846 14 8 W 5.0000 13 9 A 4.6154 12 10 D 4.6154 12 11 T 4.6154 12 12 B 3.8462 10 13 H 3.8462 10 14 I 3.8462 10 15 M 3.8462 10 16 Q 3.8462 10 17 U 3.4615 9 18 V 3.4615 9 19 Y 2.6923 7 20 F 2.3077 6 21 O 2.3077 6 22 L 1.9231 5 23 N 1.9231 5 24 X 1.9231 5 25 E 0.7692 2 26 J 0.3846 1 From what I seem to remember reading in 'Cryptanalysis' a Vig cipher will yield this sort of frequencies. They are pretty flat, compared to English. I do however know from an e-mail I received from the person doing the challenge that there is a keyword involved. I suppose I'll have to look at vig + transpositions now! Michelle |
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| catish10 | Aug 12 2007, 06:13 PM Post #7 |
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Just registered
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There is a hint given on the site which takes you a Picasso website, but I think I've worked through all the possible keywords! Any way heres the link to the challenge if anyone is interested - http://www.cipher.org.uk/index.php?p=cipher/challenge.cipher If anyone can brute-force this and come up with an answer I'd be very interested - I'm thinking this getting a bit too complicated for pencil and paper!. Michelle |
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| Revelation | Aug 12 2007, 06:52 PM Post #8 |
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Maybe picasso is the word used for the transposition. But then again, I can see words and spaces, that is not really logic for a transposition... |
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RRRREJMEEEEEPVKLWENFNVJKEEEEEAOLKAFKLXCFZAASDJXZTTTTTTTLSIOWJXMOKLAFJNNKFNXN RAGRBAQEMHIGDJVDSEOXVIYCELFHWLELJFIENXLRATALSJFSLCYTKLASJDKMHGOVOKAJDNMNUITN RRRRLJVEEEEECLYVYHNVPFTAEEEEEMWLMEIRNGLARWJAKJDFLWNTIERJMIPQWOTZEOCXKNUBNXCN RJIRPOWEANFUSNCZVDVZNMSFEKLOEPZLDKDJWSAAAAAAAOERHJCTNCKFRIMVKSOFOMKMANREWNBN RZUDRGXEEEEENFQIDVLQNCKNEEEEEDGLLLLLLAWIOSNCDARLODMTOEJXMILDFJROTKJSDNLVCZNN | |
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| catish10 | Aug 12 2007, 07:51 PM Post #9 |
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hi, Just had an e-mail come through from the person with the challenge "The cipher is multiple layered, you are in the right track! it is not easy to crack so the hint in the website will help you" this was in response to - "I thought I'd give you an update! I've been working on it for a > while, but other than working out the IC I cant get very far! I have > experimented by creating my own cipher encoded first with a keyed > substitution cipher and then encoded with Vigenere to see if there are > any patterns that show through to the IC. I managed to bring out part > of my keyword using the single letters (A and I) even though they were > encoded with the keyed sub. but that was because they were the same > (all I's even though in the final code they were actually three > different letters!) I'm struggling in how to work out what ciphers > were used. If this supposed to be something relatively easy then I > would have guessed a keyed substitution followed by something else. In > further experiments, a code first encoded with a Caeser and then > Vigenere, gives the same final cipher text as if it were encoded with > Vigenere and then a Caeser cipher. Please can you point me in the > right direction? I am relatively new to this, I have only really been > experimenting with cryptography for about a year. I can break Vigenere > ciphers quite easily, but I have no idea with multiple layered > encryption!" Dont know if this helps anyone! Perhaps I should have placed this in the challenge section! Michelle |
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| Revelation | Aug 12 2007, 08:09 PM Post #10 |
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I don't think this is right, since a substitution is not a shift. You WILL get different results if the ciphertext is longer than than key, which is the case with Vigenère. So I think we should look for a substitution and a Vigenère. The order doensn't matter.
You ask, we split  edit2: this cipher has got a name, if it is what I think it is. It is a vigenere with mixed alphabeths. I made a challenge like this for the forum a while ago. |
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RRRREJMEEEEEPVKLWENFNVJKEEEEEAOLKAFKLXCFZAASDJXZTTTTTTTLSIOWJXMOKLAFJNNKFNXN RAGRBAQEMHIGDJVDSEOXVIYCELFHWLELJFIENXLRATALSJFSLCYTKLASJDKMHGOVOKAJDNMNUITN RRRRLJVEEEEECLYVYHNVPFTAEEEEEMWLMEIRNGLARWJAKJDFLWNTIERJMIPQWOTZEOCXKNUBNXCN RJIRPOWEANFUSNCZVDVZNMSFEKLOEPZLDKDJWSAAAAAAAOERHJCTNCKFRIMVKSOFOMKMANREWNBN RZUDRGXEEEEENFQIDVLQNCKNEEEEEDGLLLLLLAWIOSNCDARLODMTOEJXMILDFJROTKJSDNLVCZNN | |
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| Donald | Aug 13 2007, 01:15 PM Post #11 |
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Which is, I believe, the result of combining regular vig with mixed alphabet mono sub. |
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| catish10 | Aug 13 2007, 02:47 PM Post #12 |
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Just registered
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Hi, So I'm looking at a Quagmire type cipher? Oh no! I've been reading through the Lanaki lessons at the moment. Can someone explain how lesson ten works where Lanaki lists all the occurances of each cipher letter and their position. Works out the difference and then creates a table of factors. I understand how to get to the factors, but I dont understand which ones are written down, and how, in the example he gives. I was playing with this just to try and confirm the period for the cipher. Michelle 
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| jdege | Aug 13 2007, 04:29 PM Post #13 |
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If it's Vig with mixed alphabets, are they all the same mixed alphabet? That is, do the multiple alphabets differ from each other by only a shift? Then one effective technique is to simply line them up. Generate frequency distributions for each of your alphabets. Then taking two alphabets at a time, shift one against the other, one character at a time, using the chi test to determine how strongly they correlate. If there's one value that shows a strong correlation, odds are the two alphabets are the same alphabet with that relative shift between them. Not all alphabet pairs will show a strong correlation, if your ciphertext is short. But some will. For the ones that do, combine them by shifting one against the other, and sum their frequency counts. The summed frequency distributions will be based on longer plaintexts, and will thus show stronger correlations. Continue doing this until you've collapsed all of the alphabets into a single alphabet. It's IC should indicate that it's a simple substitution. Apply the shifts you've discovered to the ciphertext, and you're left with a simple substitution. (This technique is, IIRC, discussed in Sinkov). |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| jdege | Aug 13 2007, 04:38 PM Post #14 |
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By Vig, I meant to include Vigs that use mixed alphabets, rather than just simple shifts. Shifts are a small subset of substitutions. And a shift plus a shift yields a shift. But a substitution plus a substitution yields a substitution, and therefore a shift plus a substitution yields a substitution. If it is a simple sub plus a shifted normal alphabet vig, the result is a shifted mixed alphabet vig. If it is a simple sub plus a shifted mixed alphabet vig, the result is still a shifted mixed alphabet vig. |
| When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. | |
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| catish10 | Aug 13 2007, 05:59 PM Post #15 |
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Hi- Thanks for your help I think I understand what your saying! So technically if I treat each possible alphabet as a mono substitution (keyed/random/shifted) I should be able to work out what each letter stands for with frequency/contact data etc. And he said on the site it wasn't any thing fancy at the moment! 
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