Yet another 10B post
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Harry I don’t understand!
How are you meant to transpose the ciphertext? Do we need to somehow decode it first and then transpose, or something else?
Sorry again! But thanks for all the help!
Solved at last… one of the hardest last round ones I think I have seen!
I think the only challenge is the transposition, not the ADFGVX part of it. But, referring to post #99353 (posted on 30th Dec), sent by Mrs-Robot, she says “keep it simple”. I can only think what that could mean to my naive brain.
If you are still stuck on the transposition, USE LOGIC
Believe I’ve figured out the first transposition (to reposition the needles). Thank you to everyone giving clues for this, they were helpful eventually!
If Harry lets this through, my clue for people who were stuck on this step is: 2, 3, 4 and 4, 5, 6. A bit evil but should help in the end.
Is the brick clues supposed to hint at the transposition aspect or the ADFGVX part that follows? I have completed the transposition, and didn’t use the brick clue, so was wondering if it’s still helpful at this point?
If anyone is searching for a clue for 10B – the encryption is quite similar to the one in 9B with an extra encryption layer. I hope I didn’t give away too much! Feel free to censor.
And now I’ve solved it. What a great experience it was doing this challenge, as well as the whole thing. Thank you to everyone here that posted clues, even if they didn’t seem helpful at first. Thanks also to everyone who set up the challenge.
My clue for those who are struggling after turning the needles into letters is – check the number of digrams you have immediately after changing the needles into letters.
Happy new year!
You might not need to transpose. When you put the text into columns of the appropriate length, which block of three are correct (contain one of each character) in all the rows?
If it’s allowed, once you have spilt the text into columns of 21, as mentioned in earlier posts, I believe each vertical pipe | only correctly matches with one pair of left and right slashes. After that, the trick is to get the right order; there are 5,040 different ways to order these trigrams once found.
Or maybe not! The staggering helps reduce that A LOT! Harry
I have:
Organised the text into rows of 21.
Only taken the 2nd and 6th block of three from each row.
Made them all into letters based on 10.8.
Assumed the first half to be column and the second half to be row (like Polybius Squares previously).
Found the most common bigrams and ranked them, and assigned them letters accordingly.
Having done these, I am just wondering whether it should be a 6×6 or 5×5 square, as 6×6 is too many, but 5×5 would be missing a letter out of ADFGVX, which all seem ot be in the text. Am I on the right track?Harry, I have done everything the newsletter has said so far (most of which I already knew) but I am still yet to find anything that gives Bigrams of the ADFGVX variety less than 27. Is there something I am missing or just not getting entirely???
-ILL
Possibly still missing something about the rerrangement of the needles? Harry
A little progress.
I’ve managed a combination that gives me valid triplets on a sample of 10 rows of 21 symbols. But that still leaves a lot of possibilities:
– swapping the three elements representing a letter
– swapping the letters within the 21-symbol brick.Hi Harry.
I have rearranged the needles so that the trigrams are all valid. However, no matter how I combine them, I never seem to get less than 36 groups of two letters. Am I missing something?
Thanks, N.
I think you might be! Harry
Does the ADFGVX grid need to be 5×5 or 6×6? Just give me a clue. Also, I still don’t understand 10.6 and 10.7, so a hint for those woudn’t be out of place.
An ADFGVX grid is, by definition, 6×6, but that doesn’t mean we have to have used all the grid entries! Harry
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