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Challenge "Double Columnar Transposition"

Challenge "Double Columnar Transposition"  

  By: admin on Oct. 12, 2011, 2:05 p.m.

Asterix and Obelix besiege Rome. They just happened to find an encrypted message but instead of the common Caesar Cipher the Romans used a new method. Can you decrypt the message?
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 Last edited by: admin on Oct. 31, 2021, 2:54 a.m., edited 1 time in total.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: jomandi on Oct. 12, 2011, 5:45 p.m.

Ich habe die Aufgabe zwar gelöst, doch leider nimmt das System die Lösung nicht an.

Wenn ich mich anmelde und die Lösung in das Lösungsfeld eingebe, dann erscheint nicht die übliche Meldung, dass die Lösung richtig (bzw. falsch) ist, sondern es erscheint wieder das Eingabefeld und sonst passiert nichts. Ich habe es mit Firefox und IE probiert. Leider beide Male ohne Erfolg…

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: Javex on Oct. 12, 2011, 8:57 p.m.

Danke für den Hinweis, der Fehler sollte jetzt behoben sein, bitte testen Sei noch einmal die Eingabe der Lösung.

english: Due to an error it was not possible to submit a solution to this challenge. This issue should now be corrected.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: be on Oct. 12, 2011, 9 p.m.

Ich habe die Aufgabe zwar gelöst, doch leider nimmt das System die Lösung nicht an.

Danke für die Meldung. Wir schauen uns das an.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: aurelie on Oct. 12, 2011, 11:38 p.m.

Der Fehler scheint behoben zu sein - Danke für den Hinweis :)

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: DarkFibre on Oct. 13, 2011, 5:43 a.m.

I wrote a program to solve this for me and I can't imagine how I would break it by hand. With a regular DCT cipher I would write out letters on strips of paper and shuffle them around, but how can you break an irregular by hand?

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: be on Oct. 13, 2011, 11 p.m.

I wrote a program to solve this for me and I can't imagine how I would break it by hand. With a regular DCT cipher I would write out letters on strips of paper and shuffle them around, but how can you break an irregular by hand?

I guess this one could be done by hand, as the hint says the passwords are two different meaningful words you can find easiliy in a German dictionary.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: Veselovský on Oct. 14, 2011, 12:11 a.m.

I guess this one could be done by hand, as the hint says the passwords are two different meaningful words you can find easiliy in a German dictionary.

But this is of no help, since ordering of letters of all German 5-letters long words represent all 120 permutations of (1,2,3,4,5). (maybe just 119, I was not able to find German word with ordering (3, 2, 1, 5, 4))
It would be help if these words represent just small portion of 120 possible permutations, which here was not the case.

But I do not say, that it cannot be solved by other method than brute force.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: DarkFibre on Oct. 14, 2011, 6:42 a.m.

So did everyone write their own program or does everyone but me know about a tool that breaks these? PM me if it's giving away the answer.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: fretty on Oct. 14, 2011, 6:51 p.m.

The fact that they are meaningful German words that appear in a dictionary makes the task no easier, there are many meaningful 5 letter words. It is just like me saying "I have chosen a number, guess what it is?" then later saying "Oh, as a hint it is between 1 and 100000000000000000000000000000000000000000000000000000".

The way I see it is that 97 letters means there must be 2 letters that "overhang" the other columns.

In decrypting the first layer of the cipher we have 5C2 = 10 possible places to put these overhanging letters.

Then after assigning these we have 120 possible permutations of the columns, meaning we have 1200 possibilities for the partially decrypted ciphertext.

Then we must again assign 2 overhanging letters in one of 10 ways and try another 120 possible permutations in order to get the plaintext.

This gives 1200^2 = 1440000 possibilities to try.

I doubt this one is easily done by hand (especially since there is no known method to check whether the partially decrypted ciphertext is correct).

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: fretty on Oct. 14, 2011, 7:10 p.m.

Am I right in thinking that every double column transposition is equivalent to a single column transposition…but with the product of the key lengths as the number of columns?

This seems to follow from group theory.

This would mean that this cipher is equivalent to a single column transposition with key length 25…still not easily breakable by hand but at least plaintext can be spotted with this.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: DarkFibre on Oct. 18, 2011, 8:53 p.m.

I don't see how it is, given the way the columns are counted off after shuffling, but a visual diagram would convince me.

Anyway, this was a cool challenge, and I look forward to the level 2 version of it, and the level 3 version which will hopefully get people started on the level X challenge.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: fretty on Oct. 18, 2011, 9:08 p.m.

See Lanaki's online lecture notes about double transposition ciphers. He gives an example of this.

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: Veselovský on Oct. 19, 2011, 8:25 p.m.

You have created a ciphertext C by simple columnar transposition with key K of plaintext P.
Then decrypting C with the same key K will produce the original plaintext P.

I have a question whether there exist a different key, say L, that when it is used for encryption of ciphertext C will produce the original plaintext P?

Hope it is understandable.

My personal feeling/intuition is that in general it does not exist.
I forgot to say, that length of K is equal to length of L, otherwise it obviously exist.

Nevermind, I have solved it :-D

Re: Challenge "Doppelwuerfel / Double Columnar Transposition  

  By: fretty on Oct. 20, 2011, 6:15 p.m.

This is a symmetric cipher, you use the same algorithm to encrypt/decrypt.


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