I have the following ciphertext that I want to decipher:

bnmkogt vgyhni uioopkm xswercvdf yuiogvbnm ghjy cftyhng tgeyjn xdrtygvc zsefx xdrgbhu tgbnmko tgbnmhyui yhnmko vbnmgyuiok

So far, what I have done:

  • Perform a frequency analysis. This suggests that no transposition cipher was used. I Also try all 26 ROT combinations for substitution cipher with no results.
  • Move to polyalphabetic ciphers. My first option is Vigenere. In that context I found that there are several trigrams , bnm, repeated and two 5-grams, bnmko, (which is probably not random and indicates that the same text is encrypted with the same part of the key).
  • I use the only 4-letter-word of the text, GHJY, and compare it with the most common four-letter words in English (assuming it is English...), with no good results.
  • I measure the distance between the 5-grams and found that is 72, which factors are 1,2,3,4,6,8,9,12,18,24,36.
  • I also ran an Index of Coincidence analysis which suggests that the key length is 7.

I think that I'm missing something. Any thoughts/ideas on how can I better address this cryptogram? What would you do? Thanks!

PS. The only clue of the cryptogram says "Manual work". Apparently, it is a classic cipher...

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  • $\begingroup$ Are you sure about the distance of the 5-grams? You need to compare the start of each 5-grams, without spaces! $\endgroup$ – Maarten Bodewes Oct 28 '18 at 5:23
  • $\begingroup$ I have removed the spaces when I did the distance count (72), but did not start in the beginning of the 5-gram (I did it at the end). Then the distance is 77, which factors are only 7 and 11. Being 7 the number I got in the Coincidence Analysis!. Excellent observation @MaartenBodewes! $\endgroup$ – rpatrick75 Oct 28 '18 at 5:56
  • $\begingroup$ Still unable to crack it. Ideas? $\endgroup$ – rpatrick75 Oct 28 '18 at 19:08
  • $\begingroup$ Well, there are likely only $7 \cdot log_2(26) \approx 7 \cdot 4.7 \approx 33$ bits of key space, so you could try and brute force it. But you should probably assume Vigenère and just trying keys and checking if all the letters in the right column check out (using 7 columns, of course) would significantly speed up the search. Remember the word boundaries. You probably don't want an overabundance of Q's and Z's :) $\endgroup$ – Maarten Bodewes Oct 28 '18 at 20:45
  • $\begingroup$ You mean that I should create columns using the letters that are ciphered with the same letter of the key? - letter 1: 1,15,22... - letter 2: 2,16,23... an so on until letter 7. In that case I will have 7 strings which presumably are ciphered with the same letter of the key. The result is the following 7 strings. bvisfbctdzrbbybo ngowynfgrsgnnhnk myoeumtetebmmnm khprigyyyfhkhmg onkcohhjgxuoyky gimvgjnnvxttuou tuxdvygxcdggivi $\endgroup$ – rpatrick75 Oct 28 '18 at 21:52