Can Anyone Break Collin152's Cipher?

[Arrogonios]

If I got the spelling right on the plaintext...
Name:
MOICNEGELIKUOP

Figure out the de-coded version if you can.
Bwahahaha!

Here's the story: Collin has actually (in his past) kissed a girl. He gave this ^ to find out the girl's name. Can anyone help?

For those of you who pride yourself on codebreaking, that evil laugh and disbelief in your abilities at the end is reason enough to do it.

[First Speaker]

Social engineering lolz.

Given our otherwise observable evidence on Collin viz. love of wind-up, mind-boggling (no, I won't deign to his phrase), it is highly likely that he has provided us with a series of random thirteen letters, and is privately laughing behind his back.

Secondly, even if it is genuine, there isn't really enough to make a pattern - thus, either he's used some really easy encoding (direct letter swap, rot-something, etc, which I'm too lazy to try), or he's using something like a one-time pad - again, laughing.

MOICNEGELIKUOP

[UncleWolf]

Do you know what kind of code? Is it Play-Faire? Is it a simple alignment code?

If it is Play-Faire, then it is a 7 letter name. If it is alignment then it is a 14 letter name. But, Collin likely did put just random letters.

[Tirian]

Secondly, even if it is genuine, there isn't really enough to make a pattern - thus, either he's used some really easy encoding (direct letter swap, rot-something, etc, which I'm too lazy to try), or he's using something like a one-time pad - again, laughing.

I was not-lazy enough to use an online applet to verify that it wasn't a caesar cipher (i.e. "rot-something"), and like you say there isn't enough ciphertext to be able to pursue any other encoding technique.

In theory, there are enough vowels that it could anagram to someone's name, but I presume you would have to know Collin pretty well in order to attack it that way. On the other hand, perhaps you would have to know Collin pretty well in order to care about who he has kissed.

[Collin152]

How very funny.
I never thought of giving random letters...
Should I provide more encrypted text, then?

[Bitzeralisis]

Yeah. Do so.

[Collin152]

When:
VOGHDAZJDAZFNI

Where:
OGKPGHCATJZFOTAFACIREOPITNFOEHHRVNJN

[RS14]

If it is alignment then it is a 14 letter name.
For the record, there are no 14 letter female first names occurring in more than 0.001% of the US population as of 1990.

[UncleWolf]

Is there a key word? Are you even going to tell us what kind of code?

I figured as much

[Collin152]

That's just silly.
If I wanted to help you out, why would I have written the code in the first place?

I want to see if it can stand up to scruitiny of the greatest midns of the world: Nerds!

[Moff Chumley]

Well, Collin is ...unusual..., so I don't think anyone whom he would be attracted to is ...usual... .

[Collin152]

Well, Collin is ...unusual..., so I don't think anyone whom he would be attracted to is ...usual... .

Attraction had nothing to do with it, chum.

[UncleWolf]

sigh, Is the codeword(if there is any) even related to the topic?
Can you at least answer this?

[streakster]

Single letter frequency analysis:
A 5
B 0
C 3
D 2
E 4
F 4
G 4
H 4
I 5
J 3
K 2
L 1
M 1
N 5
O 7
P 3
Q 0
R 2
S 0
T 3
U 1
V 2
W 0
X 0
Y 0
Z 3

Coincidence Count:
1 1
2 3
3 2
4 4
5 3
6 1
7 3
8 2
9 2
10 3
11 4
12 3
13 3
14 3
15 5
16 7
17 0
18 5
19 3

Still working. Hope this is helpful.

[Collin152]

sigh, Is the codeword(if there is any) even related to the topic?
Can you at least answer this?

If a codeword, keyowrd, or any other such word exists, it is related to the topic, has an even number of letters, and is English.

If it doesn't, it's Purple.

[Moff Chumley]

So, purple must not be related to the topic? I suppose that means that Collin's girlfriend isn't named anything color related...

[golentan]

Please first assure me that the codeword/phrase (if it exists) is shorter than the maximum characters encrypted with it at the time (for example: for the first one (encrypted fourteen letters) it would have to be a 13 letter code word or less). Otherwise you are aware that the cypher you have given us is theoretically impossible and we would have to brute force it by trying every possible word of sufficient length, then word combinations growing in to the ridiculous and requiring computational power/time greater than the total contained in the universe present past or future.

If the code phrase does not exist, say so and be done with it.

In short, your presentation of the problem proves you want to test us or taunt us. If the former, you must first give an assurance that this is breakable. If the latter, you must give us the same assurance even if though that is blatant falsehood. If you fail to do so, I will use my own "code phrase" to scramble your cypher into something mortifiyingly embarassing and simply declare I have cracked it. That's also a fun game.

[Collin152]

Please first assure me that the codeword/phrase (if it exists) is shorter than the maximum characters encrypted with it at the time (for example: for the first one (encrypted fourteen letters) it would have to be a 13 letter code word or less). Otherwise you are aware that the cypher you have given us is theoretically impossible and we would have to brute force it by trying every possible word of sufficient length, then word combinations growing in to the ridiculous and requiring computational power/time greater than the total contained in the universe present past or future.

If the code phrase does not exist, say so and be done with it.

In short, your presentation of the problem proves you want to test us or taunt us. If the former, you must first give an assurance that this is breakable. If the latter, you must give us the same assurance even if though that is blatant falsehood. If you fail to do so, I will use my own "code phrase" to scramble your cypher into something mortifiyingly embarassing and simply declare I have cracked it. That's also a fun game.

There are fewer than 12 characters in the codeword.

And I'd be quite entertained if you could find an alternative key that can make sense of the nonsense.


So, purple must not be related to the topic? I suppose that means that Collin's girlfriend isn't named anything color related...

:annoyed:
Both present-Collin and past-Collin disapprove of the use of the word girlfriend here.

[Gray Jester]

There are fewer than 12 characters in the codeword.

And I'd be quite entertained if you could find an alternative key that can make sense of the nonsense.

With any given keyword of exactly the same length as the ciphertext, it's possible to get every single possible combination of letters of that length, providing theoretically unbreakable encryption.

So, if there's a keyword, should we assume you're using playfair?

---

Guys, if he is using playfair, we'll probably need more text. Even if the keyword is only 10 letters long, we don't have enough repetitions to truly break it, particularly since the name/place/etc. is unlikely to have repeating sequences such as "the" in it.

Collin, could you give us a passage? Perhaps you could describe the experience.

[UncleWolf]

I'll start using keywords that would be the most likely chosen based on the topic and running them through playfair. I'm so happy I finally get to find a use for my skills in this stuff. :smallbiggrin:

[Thanatos 51-50]

One kiss does not a girlfriend make. Nor does it a boyfriend make.

I'm straight and have kissed a guy before. In public. His name is Dylan. I'm not ashamed. It was on a dare.
Bah all you people who care about the knowledge and not just the code-breaking fun.

[UncleWolf]

I just like a challenge. What it ultimately tells me means almost nothing to me unless it is personal.

[Collin152]

One kiss does not a girlfriend make. Nor does it a boyfriend make.

I'm straight and have kissed a guy before. In public. His name is Dylan. I'm not ashamed. It was on a dare.
Bah all you people who care about the knowledge and not just the code-breaking fun.

Yes!
Huggle!

Sides, was too young for the term girlfriend to be believeable.
Is just silly, so we took offense at the use.

We being me.

[Telonius]

The girl's name was Moicnegelikuop. Foreign exchange student from Turkmenistan.

Do I win? :smallbiggrin:

[ghost_warlock]

If a codeword, keyowrd, or any other such word exists, it is related to the topic, has an even number of letters, and is English.

If it doesn't, it's Purple.

Hm. Gem Flower's location is listed as Purple. Maybe she knows something...

Or, this could be a wolpertinger... >=)

[UncleWolf]

If you could do it without revealing too much, could you separate the words in the where part?
Its okay if you say no, but it'll be a hell of a lot harder if you don't.

[Collin152]

If you could do it without revealing too much, could you separate the words in the where part?
Its okay if you say no, but it'll be a hell of a lot harder if you don't.

I will not seperate the words in any of them, sorry.

[13_CBS]

The girl's name was Moicnegelikuop. Foreign exchange student from Turkmenistan.

Do I win? :smallbiggrin:

I...I think you might have insulted every Turkmenistani in one fell post. :smallfrown:

[Flickerdart]

Both present-Collin and past-Collin disapprove of the use of the word girlfriend here.
What about future Collin and Bizarro Collin?

[Collin152]

What about future Collin and Bizarro Collin?

Future Collin has a soulpatch, ponytail and katana. He disaproves of everything.

Bizarro Collin...
Wouldn't be available to talk. He'd be with at least three of his seventeen lady friends.

[Recaiden]

Are you using the same code for all three areas of text?

[Collin152]

Are you using the same code for all three areas of text?

Yes; I may be evil, but I'm also kinda lazy.
Otherwise, i'd use a differant code for each word, with riddles to identify codewords for each line, each of which must be tried for each word until you find the one that worked.

[Recaiden]

Yes; I may be evil, but I'm also kinda lazy.
Otherwise, i'd use a differant code for each word, with riddles to identify codewords for each line, each of which must be tried for each word until you find the one that worked.

Can we have the riddle then?

[Collin152]

Can we have the riddle then?

"What haven't I got in my pocket?"

[rankrath]

"What haven't I got in my pocket?"

your hands? hmm. so you kissed her last og, at the ge ec, and her name is gt?

[Recaiden]

It's not hands. So what other word with an even number of letters could it be? It isn't in his pocket, and is somehow related to the situation.:smallconfused:

[UncleWolf]

His cape is purple. and not in his pocket.

[Ivius]

This probably isn't much help, but I wrote a script to test out all the rot-ns and got nothing.

[Lady Tialait]

I got it! Collin kiss the most popular girl in the world!

All three of Hanson.. (http://www.youtube.com/watch?v=oCdMqpsUc2o)

[Recaiden]

I'm not sure that we can solve this, given the fact that we have such little text, and it is primarily composed of a name. Also, were the time and place ciphered with all of the text as one cipher, or each piece separately?

[Collin152]

Also, were the time and place ciphered with all of the text as one cipher, or each piece separately?

That is irrelevant.

[Signmaker]

"What haven't I got in my pocket?"

I'm going to take a lord of the rings-esque guess here and throw out "Ring" as a possibility?

[Recaiden]

It's not ring, and none of the common repeating digraphs come up with a even codeword related to the topic.

[Thanatos 51-50]

Collin is Hiro from Heros?

Or, Future Collin is, at any rate.

[InaVegt]

While I don't have the answer to his cipher, I do have one of my own for y'all to crack.

lvcwhwmupvtysesvgwkunvihbggufktzfwfsoxgnsmkefibiem gfyltvcif

Edit: Ignore the space, boards put that in.

[Ashtar]

Working on Gezina's: We have a very short corpus of text which makes it harder to work with.
Here are the frequencies: a 0, b 2, c 2, d 0, e 3, f 6, g 5, h 2, i 4, j 0, k 3, l 2, m 3, n 2, o 1, p 1, q 0, r 0, s 4, t 3, u 3, v 5, w 4, x 1, y 2, z 1

Looking through the digrams only yields one repeated digram: "vc" with a frequency of 2

Looking through the trigrams (and higher) yields no repeated frequencies.

This would point me to suspect that we are not dealing with a monoalphabetic substitution cypher. First we would Kasiski, but here we have no repeated groups to work with, so that fails. We try the Friedman test and get an IC of 0.043 which is inferior to the 0.065 of the expected english language. So we are probably dealing with a polyalphabetic scheme. We should try Hill's system, but again the only repeated digram "vc" is of little use.

Maybe more on this later if I am inspired.

[Cobra_Ikari]

It being Ina, I'd guess it's not English. :smallwink:

[InaVegt]

Oh, don't worry, both the encoded text and the key are in English.

[Recaiden]

I still think that the one repeated pairing in Gezina's and the 2 in Collins are statistically insignifigant, as they don't provide any correct keys, unless the key is something really wierd, like zymgenoid.

[Ashtar]

On Gezina's text, an IC key test gives 4.805 suggesting the key might be four or five characters. Again on such a small corpus with no repeated digraphs, it's going to be very hard.
There's a few things we could do: Rule out a monoalphabetic substitution cypher by generating a brute force attack on the text, this could be possible if we knew of a word that existed in the text.
Otherwise we must wait and hope to intercept more transmissions to gain a larger corpus of text, as long as the key and encypherment is unchanged.

I will give it another look, but there's only so much you can do with a small cyphertext.

-- or I could whimp out and go and ask Serge (http://en.wikipedia.org/wiki/Serge_Vaudenay) Vaudenay (http://lasecwww.epfl.ch/~vaudenay/) one of the professors I had.

[InaVegt]

I'll give y'all a new puzzle, with a new key. And I hope y'all will have fun deciphering it. In order to help y'all with deciphering it, I made it longer.

LGGIOBGWCEKOFBKAVZEGVHRSGZIUIIVNVVOIKDUEWVVYQLTENA TGSPMLEAJHFVRYTGPKPTOMWCXPCIQBRLEULATPIDFACBPWXEYX HDCAKEXBMYWHXCLUMPAJVNNQXUNZEPBGXVTMTQUFWUGITFBXRG PWTJAQIVVFDPHWGXVEEQUSFOKELPBPACBPREMFXHRPNKXNMQSJ AHJQVMGIBAAXCUDCIMBXIVQPQSWVWIKUWFMGRBVHTQSKWKTWUP IPMONKKBCUQICEIIISAWWBXSCXAUZTAIYMYIVPEKQWTTYXPXHO IOXNUGMZBRUWBJZSIVAPWZOHECZEYPBWFUKRMHNXGLRMHZWNHX SNRMHBSDBWKBVCERIRXVISNIQEXLLAQAAUIWLBBTLPUOFIGFAM CLOZSGZAJMPDNOBNSTNLMJGWPPENKBGESIUZRUNASIUEVVJWZL MMGZXAHAOPNVAAYXXSUXJQSLAMYEWCXETTFMYZKLOQMMGUSYLU TGWTZMVNTKASQYPLIXMJBAFOOQOYAIKNDLYWWTHGRSARSMYXVJ ZSEIGNMHQAASWMPOTZGXIQFTNPGHVQRXCPEKGHQXFRBFAMVVIV GINQZRNTIUTHVURMWELKLHMJWHVRXLTXWSGVRWGEVOVPAJVIEE IYTAOEBNKIZVRVJRSMWGNGZSALZVGBNVMYIHZHYIUZRQGBMFCF NSVAZNNIJWFWACEWMFHTCCFIVUBMJPLAKYXMLQFWEASEMIFTBY WVMMGNIWKPWZGXZFXQHRFRVKNIKXFHGFYIISEWJTEXZVLCILVI FTSCJEIACAOWVXIGLNIDDAENFWBWJATVWRRFMEPYTVTWIJYFMC MCIENGEMBQENEBAYBWTXSYJMLNARMWHOIKGYXZTYIAZSFCCCGY KPADAINRGHOMIFILQSDCVUGFGXMJHDAEZYWIMDUNPIFVQMACHC GLVEYLLIKKKLUPQFUEGMJLSHUWMMVXYQSRUCEHZVQHTHHYWZOS IGDBIBFVTAXUALHWTCRUSAWVTHLSICSLCLMHZTZSFYAQOYHQVU ATKCVSWFETYFBNLGIRMKXPWRZHBAQRRZNUJOAYPYCYHWIUQYJB GNMTYRKZTKFIPYXWXXTTJYJJVBVRYIXWHVUHEVRZEOBVKEQIMN UOQWUAGEQRZTIMLETURGVKGXAYGFEUHNDYBRUJLAEUWQFEHUQF ITLSWYIKBNLLVSSVRMXVVDGCSAEZHUWMZZEQPFMHQIDUODBUPE DYQJXFVANWCJLUGRTXYOCLTCGXYEHKWXFCPBIEJCFLSYMRAGUA DJXQMWSFIOCQIXLJJUVCBXWGOHNSJQCFCAREMEOFOWVOBRHEII ZPVVQJSXKDGTIGEYEJOKHDDCZFPQTYSUTNDEICVLJSZAHFEITA IJFGNLNNTRLQLDDGVIZNKMBTEGLMFQFDNGNZRZCEXVBJALQNGB GMHKYDUJQZIDUBXTUNTPZLVVUMNFCLMWWTALBEAWFGGRAHEKII LBLLBIQHJTLTTOVICYVGYXKXBMIMSPMGGSGJPAHRWLGRWGIIUI ZSFBUWVXLVSAANXUHTFJDKAZBTLRJNGHGOYISYUPWJBAWCULDI OXRUWATGDWUUSAFECMIYXSULAFBRFVRKTNISFYDYEXNOAGQMTO PABNORGRWAVMGGNGHTFVYLSRHHIJRGNIZNIOOPIQABCIFCDANK FOPVPAWIBEHZMKOOBGSHURFEOVBIZLGTSGQGFWWUXWAIAVAILZ PILQCDEIBTVNRILCDETXZUSVIMREMRXYVWEDSHIONSYWGIEQLE OWZMLBUASQSHCMKAQGEEMILFOVHGAZZGXKZGMHIITFBQGVYERX AFSXBTEOAXXICRMBBXVBVPLHDZRLAWMOBIMEBPQAENQGPYSVVH BUGZEIHKYIAMYMXUYWGZQTILPVLETSKSEBAEEIKSAGQJBMEYXH DMTGQWLAFRRSLLHCSIHNIKQDTMRAAMDHSWWPIXRLEZYEFTUJND NSSDHWSKYFLVKVGKRGGZGJTVJBWBYQWNIASSXYSTGYUXMGCKLS JITEWWZQDWIEP

As with the previous, any whitespace is inserted by the forum.

[Ashtar]

There are 2 ";" and a "2" inside. Can I ask if they were present during encypherment and unchanged or were they skipped ? You don't need to answer, we could simply run both cases, but it would make things simpler.

Friedman test gives an IC of ~0.04 which is close to the 0.038 of "natural" random, so we can assume it is not a monoalphabetic substitution cypher (unless it's obfuscated - padded).

Kasiski and di-/tri-graph will follow.

[InaVegt]

I forgot to remove them from my source text. They are removed now, so if you haven't saved the text, bad luck.

[curtis]

I'm proud to say that it was I that provoked Collin into unleashing this monstrosity on the world.

[Collin152]

I'm proud to say that it was I that provoked Collin into unleashing this monstrosity on the world.

Strictly speaking, it was Arrogonios, who was also pestering me in the real world. Given how we know each other and all.

(By the way? He's awesome and adorable.)

[Ashtar]

Did some more testing on the Gezina corpus, it seems a polyalphabetic substitution cypher should be ruled out due to IC on the columns (for keylengths 1 to 15). It could still be a polygraphic cypher. I'll know more when I start doing the repeated digraph, trigraph analysis.

Sadly, I've got too much work to really give this any time (- had less than 3 hours sleep last night - working). Anyone else still working on it?

For the Collin152 corpus, it might be possible to decrypt it if he's using a monoaphabetic substitution cyper and if we get the correct word spacing - of course for that we need his help. Using that, we would brute force the second and third messages doing a dictionnary attack and checking possibles. The "When" message might be the most interesting since we could restrict the dictionary to time/year/month related and try some mappings.

Again, this is quite hard since I don't know which type of cypher I'm working against, I've got no access to plain-cypher text pairs and in the case of the Collin152 corpus we have a very small data.

[Fiery Diamond]

Why were you positive "ring" wasn't it again?

[InaVegt]

Ashtar: For mine?

Vigenere cypher with a randomly generated key of length longer than the plain text.

[Ashtar]

Which basically means that without a plaintext - cyphertext attack, it is more or less equivalent to a one time pad and thus uncrackable without spending an exponential amount of time.

I spent my time assuming that it was a feasible problem due to this:


The primary weakness of the Vigenère cipher is the repeating nature of its key. If a cryptanalyst correctly guesses the key's length, then the cipher text can be treated as interwoven Caesar ciphers, which individually are easily broken. The Kasiski and Friedman tests can help determine the key length.

But your answer brings me to this:

The running key variant of the Vigenère cipher was also considered unbreakable at one time. This version uses as the key a block of text as long as the plaintext. Since the key is as long as the message the Friedman and Kasiski tests no longer work (the key is not repeated). In 1920, Friedman was the first to discover this variant's weaknesses. The problem with the running key Vigenère cipher is that the cryptanalyst has statistical information about the key (assuming that the block of text is in a known language) and that information will be reflected in the ciphertext.

If using a key which is truly random, is at least as long as the encrypted message and is used only once, the Vigenère cipher is theoretically unbreakable. However, in this case it is the key, not the cipher, which provides cryptographic strength and such systems are properly referred to collectively as one time pad systems, irrespective of which ciphers are employed.

Added emphasis to the second paragraph. So basically I was attempting analysis on a one time pad. I'd have more chance using social engineering on the person from whom the message originated... Can I get you drunk and get the message out of you?
:smallwink:

Well thanks for that, that clears it up. I don't need to spend any more time on it.

For anyone who is interested in cryptography I recommend "Cryptological Mathematics, Robert Edward Lewand, Mathematical Ass of America".

--
Nuts!