Challenge "ADFGVX — Part 1" ¶
By: admin on Aug. 15, 2011, 5:12 p.m.
A small ciphertext encrypted with the ADFGVX cipher, which has been used during World War I, is given. Are you able to reveal the plaintext message?
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By: admin on Aug. 15, 2011, 5:12 p.m.
A small ciphertext encrypted with the ADFGVX cipher, which has been used during World War I, is given. Are you able to reveal the plaintext message?
Read more...
By: vosi on Nov. 22, 2011, 5:39 p.m.
I have the same problem as Dark Fibre in this challenge. Must be the answer wrote in CAPITAL letters with spaces between the words and without punctation???
Hi,
yes, the solution must be sent in uppercase characters with spaces. No punctuation because ADFGVX has only A-Z and 0-9 in its alphabet.
By: bgr on June 27, 2013, 11:14 a.m.
Dear community,
I am stuck now!
I built all 24 possible transpositions of the cipher text and generated the related bigrams.
Then I calculated the bigram frequency. Due to the fact that the letter frequency must be E > I > N, I reduced the possible solution to 8.
For these 8 possible solutions I generated the ADFGVX matrices. I eliminated again the matrices, where e.g. the 'N' is represented by the bigram 'AD' and other impossible formations.
Surprisingly only 1 matrix is left now and I already have three letters of the code word. I tried all combinations with the missing ones, but no german words appears in the plain text.
And I cannot find a mistake in my Excel spreadsheet and have no clue now how to proceed. Any hint where I understood something wrong?
bgr
By: bgr on June 29, 2013, 12:25 a.m.
Hi Bernhard,
have you doublechecked your bigram substitutions? Maybe there is a mistake in the mapping of bigram frequencies back to the characeristic letter frequency. The codeword used as substitution key is a standard german word.
Good evening,
it took some time, but I doublechecked my self-made generation of the bigrams, looks fine. Calculation of the bigram frequency looks fine, too. Did the replacement of bigrams to letters manually, but looks fine, too, for all 24 theoretical possible solutions. After eliminating the solutions where E > I > N is not valid, I only have 8 possible solutions left. Created the matrices for these 8 by adding E, I and N. At the end I have one matrix left which makes sense. But no chance to find the fourth missing letter of the code word. Checked with Cryptool 2, same result.
Will try to find a solution to automize the whole calculation somehow to get rid of some manual steps. Hopefully it's an issue of miscalculation instead of misunderstanding!
Bernhard