Yet another 10B post

[Olisa]

I think the common confusion amongst those of us who haven’t gotten 10B yet is what to do with the trigrams or if we should have trigrams at all. The cipher 10.8 hints is pretty obvious but how we get to that point isn’t. As well as what 10.5 – 10.7 even mean. Any insight on one of these would be greatly appreciated.

[TheCipherCrackers]

We have decrypted 10A, but are still not making much headway on 10B. We have decided to write down the things we know / have observed in the hope that Harry might allow these jumbled thoughts (which may or may not be useful!) to be shared and that these might trigger some new discoveries…

1. we understand where case files 10.2 and 10.5 come from but are unsure what the others mean
2. we worked out that the smallest block size where there is always an equal amount of slashes, backslashes and pipes, is 21…and 42 is a multiple of this which ties in with Cribbage’s kind hint above!
3. we think that each brick in the wall (shown in casefiles 10.6 & 10.7) could represent 3 needles (because that’s what 10.7 shows!) and each row of bricks could then consist of:
—> 7 needles with staggering pattern: 3-3-1 / 2-3-2 / 1-3-3 / 3-3-1 (all add up to 7 needles per row)
—> 14 needles then we’d have pattern 6-6-2 / 4-6-4 / 2-6-6 / 6-6-2…(adding up to 14)
….
—> or perhaps most promisingly…42 needles and then we’d have 18-18-6 / 12-18-12 / 6-18-18 / 18-18-6 as the pattern
4. but staggering the original ciphertext and then still reading it left-to-right and top-to-bottom doesn’t change the order of anything and we then still end up with lots of blocks of 3 that would be invalid configurations of the 3-needle telegraph (e.g. ‘\\\’ which doesn’t point to any single letter)
5. so we think we need to “re-order” characters in the original ciphertext in some manner before doing anything else…or read them in a different way…which we have not yet worked out!
6. if we read the ciphertext *without* reordering (in blocks of 3), and translate all blocks that we can based on the 3-needle telegraph shown in 10.8, then we notice that when we group these into rows of 7 blocks of 3 (21 ciphertext chars per row), the 2nd and 6th columns are ALWAYS valid configs that point to one of the letters A/D/F/G/X/V (which are those shown on casefile 10.8). If we have 42 ciphertext chars per row then it’s the 2nd / 6th / 8th / 12th columns where we get 100% translation

Are we making any sense at all!?!!

This is really excellent, which is why I held it back until Jan 2nd. You are DEFINITELY on the right track. What does Case File 10.8 suggest about the type of cipher this might be, and how does that fit with your thinking above? Well done, keep going, Harry

[Sprinter390]

You can split the text into smaller sub-groups that have the same number of / \ and |. Appears that you need to transpose these groups to obtain something meaningful but have no ideas on how to actually do this.

[someone]

I found the correct transposition key to get 1 of each symbol in each chunk of 3 and do not know what to do from there. My tip is that anyone who can do python coding should create a code that can solve transposition ciphers based on a key you give, and chunk size; i’ve done this for all the other challenges – i created 1 script that can decrypt and encrypt Caeser shift, Substitution, Transposition, Standard Vigenere, Autokey Vigenere, Nihilist, Polybious sqaure and auto-solve for Caeser and Substitution. I recommend also creating tools for finding chunks in certain locations, frequency analysis, permutating, and converting keywords to grids and alphabet.

[_madness_]

All right, children, here it is: The point of the bricks is to say that the size of the blocks for shuffling is different from the size of the blocks that specify letters.

[someone]

Harry could you please give a clue on the type of cipher(i think it might be ADFGVX) as that would help at least a bit. Also i was wondering if you hold any other cipher challenges like this over the course of the year as this is a great challenge and i wish it happened more frequently.

I think I already did! Thanks for the kind words. I am afraid we usually only run the one competition a year. It takes quite a lot of effort and time, and work will start on the next one more or less as soon as this one ends. Harry

[DrPrasad_13]

Harry, what do you mean by the trigrams? Do all of them need all 3 symbols

[EllipsisEllipsis]

I noticed that if the cipher is divided into equal sections of 3 * 7n (such as 21, 42, 63, etc.). The number of /, |, and \ are always equal in each section. Perhaps there is a certain way to transpose the special characters in sections so that 21 special characters represent three needles (7 special characters represent one needle) and bring us closer to the answer.

[eel-eel]

I’m not sure how much of this will be let through because its quite revealing so feel free to censor as much as you want Harry. but if you continue down that path you can whittle it down to chunks of 21 which contain 7 of each character. I believe this is what the case file with the brick wall and the one about counting is hinting at. The last hint is an obvious pointer towards the ADFGVX cipher which I should have seen earlier because the cipher is famous for having only 6 characters and the three needle telegraph can only output 6 different characters. The part I’m really stuck on is figuring out how to extract the 7 letters from the blocks of 21 or any letters from any sized chunks for that matter because some of the needle arrangements make no sense. I believe case file 10.7 is trying to hint at how to figure this out but I am at a loss trying to find the meaning behind it. That’s my take on 10B, I don’t think I am far from the solution but but I am really getting tripped up by one bit.

As you can see, I let all of this through, but with a delay as I think it might be a bit too helpful for today! (29th Dec) Harry

[Sam]

Hi,
I have successfully transposed and translated the characters so now I have a text composed of a,d,f,g,v,x, but now I have no idea what to do with it if it is a modification of the cipher. Any hints for this point?

 

There will be, yes! We are publishing a “how to break it” guide over the next week so everyone who follows along should have a fair chance of cracking 10B. The hint you are looking for will be towards the end of that, so do keep thinking about it and see if you can get there first! Good luck, Harry

[ILL]

Can anyone give a hint, that includes you Harry, for the first key of the [Tabular Transposition]. There are +60480 combinations I know of that would give a valid chain of characters, but all I have tried have not given bigrams less than 27 when transformed into [ADFGVX] via the 3 needle telegraph. For clarity this is the [21] long key that is first use to transpose the characters into 1:1:1 [7:7:7]!

I am starting to give up…

-ILL

The question is, s that the end of the beginning, or the beginning of the end for you? As mentioned in another thread, there will be some fairly explicit guidance on how to break this cipher released in stages in the News over the next week (though thinking about it, I am not sure I mentioned in the other post that that is where to find the clues, whoops!) So don’t give up, just let it mull for a bit. Harry

[ILL]

Okay, after some calculations there are 1410877440 possible keys that can still produce valid arrangements for the transposition key. However, just a bit more than 1 gives the bigram count we need [less that 27] for the correct solution. I’ve found this number through this code:

[Edited by Harry]

Then I coded a script to go through every key for Tabular Transposition (I won’t reveal it yet, but don’t threat it still doesn’t work), turned each chain of three into a number 1-6 and checked that the total number of bigrams (e.g. AX, DV, FX…) was less than 27. So far absolutely nothing, and I’ve only gotten 50762 keys into that massive number.

Am I on the right track? Or am I just blind to the answer that has been staring me in the face since last year?

-ILL

It is a great idea, but I think you are not making the most of the brick clues! (Sorry we mostly don’t post code so we don’t end up with everyone copying everyone else without writing their own. In this case we very much do like competitors (re)inventing the wheel for themselves! Harry

[Olive-Tree3]

Wait so Harry do we have to translate all of the trigrams in the ciphertext into adfgvx? I’ve split the text into blocks of 3 characters and then rows [Edite by Harry] (if that makes sense – all these ‘chars’, ‘blocks’ and ‘chunks’ are so confusing) and noticed that the 2nd and 6th columns always have a ‘v’ in them which point to a letter… however I’m stuck on what to do from here as, assuming all the trigrams need to be converted, some of them still don’t point at a letter. Do we have to look at the text from a different angle perhaps? Am I on the right track?

I think the forthcoming hints in the News over the next week will answer your question (assuming you don’t actually get there for yourself before then. I think you might!) Good luck, Harry

[Statto]

Humph… tricky customer this, and like doing the cryptic crossword when you cannot get inside the setters mind, it is still putting up resistance, despite having had the \|/ in each trigram for some time 🙁

I hope there is only one transposition of the sets [Edited by Harry] that leads to a full set of suitable trigrams – if so, I have it! – but not yet got the later keyword and changes required to bring the bigram count down to 26 or less, and I am not an expert programmer of any kind at all, so much donkey work still needed 🙁

Good to know that the workings of my mind are still a mystery! Harry

[Cipherman-into_the_dark_web]

For the transposition at the beginning is the key length [edited by Harry]? or am i not allowed to know?

You are intended to know! But that is not to say whether your suggestion is right or not. More will become apparent in a series of hints to be published from Monday in the News! Harry

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