Yet another 10B post

[LogicPi]

I am still slightly lost!

I have rearranged the groups of 21 so I have 7 groups of 3 unique symbols (/\|)

Then, I converted each group of 3 into a new letter (ADFGVX) based on the compass.

If you take the bigrams of consecutive groups, there are 36 different combinations (but only 26 letters).

Thus, I rearranged the groups, taking 7 pairs from each group of 14 (2 lines), and each of those pairs only have roughly 23 (<= 26)combinations throughout the text.

I then tried applying a Beaufort cipher with key of length 7, but with no luck.

Am I on the right track? Have I made any mistakes? Any help would be greatly appreciated.

Not really I am afraid, though I love the idea. Breaking things into groups of 21 as you did at the start is great. But then you need to work in each block of 21 to try to get things into 3-needle telegraph code. Hopefully the last hint few hints we published in the news will have helped to see how. Harry

[eel-eel]

I’ve definitely gotten the transposition because somebody said there is only one which works but I fear that the resulting ADFVGX code is hard to crack at best. I wonder if anything else has been done to it or if I just have to search for keys?

Once you have got things into the ADFGVX form it really should be pretty straightforward to do the last step. You might be overthinking it! Harry

[someone]

After finally solving 10B, my clue is that you should look for patterns in chunks of 7 based on the bricks hint.

Excellent advice! Harry

[Meera]

Harry this is urgent, i need your help!

I didn’t quite understand your hint from the NEWS today, so could you help explaining it? If you can’t do that, can you give me some hints as to what you are trying to say? I have no clue what you meant even after reading it like 5 times.

I don’t know if you will let this through but it would be EXTREMELY helpful if you did, it will at least give us a chance of getting somewhere with Part B? There is literally only 2 more days!

OK, how’s this: having written the cipher text in rows of length 7 you can split it again into blocks of three rows, each giving a 3 x 7 grid of cipher text characters. Hence the appearance of the number 21 in lots of the discussions here. You need to permute the columns of the block like you would in a column transposition cipher so that when you have finished that the 21 characters in a block can be read off as seven triplets each containing exactly one of each of the characters \,| and /. The description in the news bulletin tells you how you might go about working out which transposition of the seven columns in the 3×7 block does the job. Try reading it with that in mind and see if it helps. (All the 3×7 blocks use the same transposition by the way.) Harry

[eel-eel]

After more toil I really can’t figure out what is going wrong. The only way I can think that I am getting this wrong is by having very coincidental errors in my code which would be embarrassing or finding a key that also happens to decrypt to a valid sequence I’ll post my code and I know how we feel about code on the forum and it does contain the key but I think it’s quite interesting how it outputs a valid sequence of telegraph needles which somehow when in ADFGVX form just doesn’t work for me.

As you say, we prefer not to post code, so I have taken that out below, but I will say that you have the right transposition, so it should just be the substitution step you need! Good luck, Harry

[Edited by Harry]

[_madness_]

@F6….Frenchy,

This is from a *different* ciphertext that was encrypted with the same cipher but *different* key.
You can figure out what these things mean.

triples that belong together are on a line together:

0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
*  *              *
      *     *  *
         *                 *     *
                     *        *                 *
                        *           *  *
                                          *        *        *
                                             *        *  *

fold over into blocks of size 7 and rewrite it sideways:
0 *  * * 
1 *   * *
2  ***   
3   ** * 
4  **   *
5  *  * *
6 *   ** 

line them up so that triples are contiguous (with wrap-around):
0 *  * * 
6 *   ** 
1 *   * *
5  *  * *
4  **   *
2  ***   
3   ** * 

this leaves 14 possibilities: 7 ways to choose first thing, and 2 ways to decide forward or backward

[Cipherkings]

I have finally finished the steeplechase/marathon of 10B, and I don’t get how the solution is head-slappingly simple. (Perhaps this is my own inexperience.) A clue, if not too revealing: once you have come up with the proper substitution, as described in previous posts, please look up what ‘ADFGVX’ actually is on Google, and then do a simpler version of it. The letters in the Case File diagram could have been any random letters, or even 1,2,3,4,5,6, but ‘ADFGVX’ was very helpful.

Anyway, thanks to the Cipher Challenge team from Cipherkings, and have a happy New Year!

[TheGryphonSchool]

Thank you to all involved, I have really enjoyed the challenge and just cracked 10B.
Simple in the end once the 7 columns of pairings of letters had fewer than 26 combinations. Then just needed to tweak the original order of three characters /\| so that the columns of unique pairings of ADGHVX were the same.
Looking forward to next year already!

[TheCipherCrackers]

Glad we finally cracked this one last night. 20pts in the bag but the satisfaction of having completed all challenges prior to final deadlines easily tops the points, even if we did need all the help we could get on 10B!

Hindsight is always 20:20 but it was actually fairly simple and we had it in our grasps quite a few days ago but couldn’t quite get over the line til now.

One rabbit hole we went down, which should be now be clear is wrong, is to label each occurrence of a {/|\} based on the number of times it’s appeared in the ciphertext up to that point and then pair all the characters with the same label. That guarantees valid triplets which all translate via the ADFGVX square but you get 36 bigrams which isn’t what you want. Probably obvious at this point given the hints, but transposing columns is the magic. We did it via a relatively quick brute force approach but Harry’s logical approach to work out the column order is easier to follow.

Thank you to Harry and the team for putting together a challenging and enjoyable set of ciphers. We’ve loved taking part. And thanks also to all those who helped with tips and encouragement on the forums!

Good luck to everyone still trying to crack this ahead of the final deadline!

Thanks for your kind words, glad you made it over the line! Harry

[The_Bombe]

Is it a Vignère?

Nope! Did you try using the hint in  Case File 10.8? A Google search will point you at the right cipher, or at least a cipher on which 10B was based. There is also quite a lot of info in the news items this week. Still time to get it done! Good luck, Harry

[The_Bombe]

Is the first half of the text columns and second half rows like before?

No, but that is a nice idea. Harry

[HJ19OPS]

Hi
I have transposed the columns to get valid trigrams using the key ([Edited by Harry]) – is that right? But I end up with 36 groups of 6 and not 26 groups. What am I doing wrong?!

[HJ19OPS]

Harry – do I need to redo the transposition part? I have a feeling I do…..
Really grateful for any advice!

[The_Bombe]

When will we start having hints about mistakes for submissions for 10B?

Oh, good point. Not sure they will help much, as you either tend to get it or not, but I could switch that back on now! Harry

[Olive-Tree3]

Hi Harry!

I have transposed everything into ADFGVX but I am still utterly lost on the key… also do we have to use numbers as there are 36 spaces and only 26 characters in the alphabet so please helpppppp

We really want to get this one

If you have the right transposition then after doing that you won’t have more than 26 sextuplets, so I think you might need to go back and try again. Harry

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