Yet another 10B post
- This topic has 12 replies, 53 voices, and was last updated 1 week, 5 days ago by .
-
Posts
-
A grouping of twenty one characters provides an equal number; of seven of each of the characters \,| and / in any one group. With seven groups of three characters, in each group, that’s six possible orderings, for each group, and (7 x 6), or forty-two ways of arranging all twenty one characters, while still maintaining equal numbers of each character in each grouping. I see the pattern looks symmetrical, which may reduce the possibilities, but that is just a hope.
Given that the telegraph cipher just provides the coordinates for the ADFGVX cipher, there seems limited scope for frequency analysis.
Well done to anyone who has solved 10B. Painvin and anyone else continuing to work through this gets my admiration and respect. I hope you get to a solution, but remember, it’s about the journey, just as much as the destination. I hope you learn something useful on the way.
Harry, feel free to censor this, but does it have anything to do with the ADFGVX cipher even though it was made in 1918?
That is quite possible. If we are too pernickety about historical ciphers we would be restricting ourselves too much and making it far too easy for you! In any case, it could be to do with the ADFGVX cipher without actually being the ADFGVX cipher – in this case it is actually very slightly weaker! Hope you enjoy figuring out why. Harry
So…been round the houses too many times on this one!
But, have transposed the original cipher to get /|\ in each group of three using 21 columns.
Then sequenced in pairs to get fewer than 36 different combinations when grouped in 7 columns of 2 letters, but there are 7 different alphabets used.
They all look like simple substitution ciphers but I can’t get a decent massage so far…is this on the right track?Interesting! Harry
Happy New Year!!! – apologies if I give away too much at this stage!
Has anyone thought about writing out the rows on top of each other?
Perchance comparing the rows (in search) for triplets could be…enlightening?
I am on the exact same point as you, and the transposition part is what bugging me. I’ve tried all sorts of transpositions and found there are +60480 possible Tabular Transpositions. However, the one of them that leads to <27 bigrams is the ONE thing that is stopping me from breaking this cipher! The [ADFGVX] part is incredibly easy now [As the equivalent is a 6×6 Polybius]. ITS THE TRANSPOSITION! I DONT UNDERSTAND!!!!
I am going mad
-ILL
Hey Harry,
I understand what to do but am stuck on the second part (if you know what I mean)… can you PLEASE try and convey in some sense the key? I have no idea what it might be because I don’t think 10.8 gives as much clue about it… Thanks.
The key to that is found by the standard method of frequency analysis. Hope that helps, Harry
Is there actually any combinations in chunks of 21 that could have < 27 frequencies when put into chunks of 6 after transposition, because i created a code and have checked for errors multiple times in many ways yet it cannot find a combination that works. Could the logic i am using be incorrect as i found a key that transposes each chunk of 21 to make it so that they have 1 of each symbol in each chunk of 3 and i am now transposing chunks of 3 in chunks of 21 to see if any combinations work. I have a feeling that i have to change the original key that is 21 long but that would yield 21! possibilities – far too many to compute. I also still think that if i knew where the sequences of numbers from 10.5 came from i would understand the decryption progress; Harry you have told me to look very closely at my results when counting but i have no clue what to count.
Dastardly cypher! Even when you can create the trigrams to give ADFGVX, of course, each of the trigrams can be ordered any of 6 ways. Much respect to your devilish ways Harry!
The hint progression felt a bit off in this challenge. We figured out pretty quickly that we needed to use 21-symbol chunks, and 10.2 backed that up. Then 10.4 made it obvious we were dealing with ADFGVX cipher.
The real challenge is how to convert the initial cipher into telegraph format – but weirdly, there seem to be zero hints about that part. It was frustrating waiting for hints after 10.4 only to see them repeating all the easy parts we already knew while skipping over the actual tricky bit 😿
(Feel free to censor parts of this since they are most likely too revealing)
@ILL i have the exact same problem but seeing as their is so many combinations – surely it’s not going to require a guess, i believe they may have hinted at it in the bricks hints but am not sure yet.
10B Definitely a disturber of the peace of mind ! ! ! Ha ha haaaar (Maddening Laugh)
After working practically every day and night this holiday, including christmas, i have finally solved 10B.
If you want a clue for the first stage, FACTORS
Good luck if you have not yet solved it, and very well done if you have (especially early on)Hey everyone,
This is going to sound like a rlly stupid question, but could someone explain what they mean by “blocks” and “transposing the trigrams”?
I am very sorry cuz this sounds rlly stupid, I am just trying to understand everyone’s comments!
Thank you so much!
Dear Everyone
it is a slightly easier version of the cipher that 10.8 points to. I was wondering if it was easier because the polybius square was smaller (5×5) not 6×6?
Am i way off?
congratulations to anyone who solved it because i am making no headwayAlso, some people are saying that we aren’t meant to divide the ciphertext into groups of 3? Is that true or do we divide into groups of 3 and then transpose the ciphertext so that we get valid combinations?
Thanks!
- You must be logged in to reply to this topic.