Reply To: Challenge 10B
This was posted on 21/12/25, but for obvious reasons we held it back. We really like it though. It is a fantastic record of the thinking behind a break and we thought it was worth showing you all. Harry
Obviously don’t show this but Harry look at the suffering I’m going through thinking of ideas (these are my notes): Characters that appear:
A C D H J K Q S X
2 3 4 5 6 7 8 9
|
\ /
Frequencies:
Chunk: S, Amount: 1281, Percentage: 0.19281
Chunk: C, Amount: 859, Percentage: 0.12929
Chunk: H, Amount: 638, Percentage: 0.09603
Chunk: D, Amount: 544, Percentage: 0.08188
Chunk: J, Amount: 346, Percentage: 0.05208
Chunk: K, Amount: 300, Percentage: 0.04515
Chunk: 2, Amount: 298, Percentage: 0.04485
Chunk: X, Amount: 297, Percentage: 0.0447
Chunk: 8, Amount: 276, Percentage: 0.04154
Chunk: 5, Amount: 261, Percentage: 0.03928
Chunk: 9, Amount: 260, Percentage: 0.03913
Chunk: Q, Amount: 246, Percentage: 0.03703
Chunk: 7, Amount: 240, Percentage: 0.03612
Chunk: A, Amount: 240, Percentage: 0.03612
Chunk: 3, Amount: 206, Percentage: 0.03101
Chunk: 4, Amount: 188, Percentage: 0.0283
Chunk: 6, Amount: 164, Percentage: 0.02468
Each chunk of 4 ends in S or H
First 20 chunks of 4: 7CXS 3H6S 7CKS XDAS 2CKS 8D4S 9C2S 7DXS 3D9S 7C2S 6HJS 9C2S 3D7S 8CQH ACQS AH5S 9DJH 7D3S 8D5S 7CQH
Possibility: DE AR DO DG E_
If you look at chunk 1 and 3, you’ll notice they both begin with ‘7C’. Now if you look at the possibility written underneath the chunks, pair 1 and 3 begin with ‘D’. So maybe that’s a meaningful correlation.
Surely this idea wouldn’t work then as it means each pair maps to a letter but there are 52 unique pairs (not 26) – however 52 is exactly double 26 so maybe duplicates.
This could be supported by a 4×13 (52 possibilities) grid with the alphabet twice:
1 2 3 4 5 6 7 8 9 10 11 12 13
1 A B C D E F G H I J K L M
2 N O P Q R S T U V W X Y Z
3 A B C D E F G H I J K L M
4 N O P Q R S T U V W X Y Z
We could also do this plus playfair rules (including modified rules like from 9B).
I could not get either of the above theories to work
If we take the first half of each chunk of 4 to get pairs, there are 36 unique pairs (even though theoretically there are 39 possibilities) (this would fit in a 6×6 grid), and if we do it with the second half of each chunk, there are 16 unique pairs (even though theoretically there are 26 possibilities) (this would fit in a 4×4 grid). Maybe first halves and second halves have different decryption grids – also no pairs from the first half ever appear in the second half. This idea would now support the idea proposed a few lines above. If this theory is correct we could lay out decryption grids like this:
6×6 grid (for first half pairs):
1 2 3 4 5 6
1 A B C D E F
2 G H I J K L
3 M N O P Q R
4 S T U V W X
5 Y Z A B C D
6 E F G H I J
4×4 grid (for second half pairs):
1 2 3 4
1 K L M N
2 O P Q R
3 S T U V
4 W X Y Z
Since the 6×6 grid contains 10 more positions that the 4×4 grid and the 4×4 grid is 10 positions away from 26 (length of alphabet), perhaps the missing 10 are put into the first grid as the first grid is 26 (length of alphabet) + 10 extra positions. Since second half pairs also always end with S or H, maybe they decide which grid to find the letter in.
I could not get the above theory to work
Big update on above theory: I noticed from the characters that appear in each index of chunks of 4 from below, the second has ‘H’, which is also in index 4. I ignored this at first however I just did frequency on pairs in the first halves of chunks that end in ‘H’, and there’s exactly 10! 26 + 10 = 36, a 6×6 grid, and I said earlier maybe 10 letters missing from the second 4×4 grid could be the extras in the 6×6 grid, and this proves it!
This idea could also be represented by using the 4×13 grid from earlier but coordinates used differently:
7 3 X 2 8 9 6 A J Q 5 K 4
C A B C D E F G H I J K L M
D N O P Q R S T U V W X Y Z
S A B C D E F G H I J K L M
H N O P Q R S T U V W X Y Z
When in chunks of 4:
first characters in chunk can be [‘7’, ‘3’, ‘X’, ‘2’, ‘8’, ‘9’, ‘6’, ‘A’, ‘J’, ‘Q’, ‘5’, ‘K’, ‘4’]
second characters in chunk can be [‘C’, ‘H’, ‘D’]
third characters in chunk can be [‘7’, ‘3’, ‘X’, ‘2’, ‘8’, ‘9’, ‘6’, ‘A’, ‘J’, ‘Q’, ‘5’, ‘K’, ‘4’]
fourth characters in chunk can be [‘S’, ‘H’]
Possibly coordinate system using chunks of 4?
There are 283 unique chunks of 4 and that is close to the amount of unique pairs you’d see in normal text perhaps chunks of 4 substitute to pairs
Unique amount of pairs is 52
Unique amount of chunks of 4 is 283
Length of text is 6644 (perfectly divisible by 2 and 4)
Harry am i close or are these all coincidences or purposeful red herrings?