The official Challenge 9B hints thread

[The_Code_Less_Traveled]

Is [please delete from here on, leaving the answer (if allowed) of course] the over-encrypted keyword {55, 14, 23, 45, 51, 55, 44}?

[Harry]

If you are still stuck on Challenge 9B, then you might want to read the brand new BOSS Technical Note 008 on Nihilist ciphers and how to break them. You will find it in the BOSS Library. It is by Madness and is an expanded version of Unit 80 in his fantastic codebreaking book, which you will also find there (BOSS Technical Note 004). Happy reading! Harry

[Blarthox_]

The Nihilist cipher is neat, but that’s not how I solved this week’s puzzle. There’s more than one way to relieve a mammal of its pelt.

[The_Pi-thon_Person]

Please, ive been working on it solidly all week. Can you at least say whether ive got the right key length: is the key length 7?

[Yes I can, and yes it is. At least, one of them is. There is a keyword for the repeating pattern on the ciphertext and another one to fill the grid. Hope that helps. Harry]

[madness]

Keep in mind that there is an error in Tech Note 008: if there is a carry digit in the addition, then there can be SIX different digits in the ten’s place for each column.

[Thanks madness, let me know if you edit it. Apologies for delaying the publication. Harry]

[The_Pi-thon_Person]

I finally did it! @Madness you are a genius. Seriously! Your book helped me tremendously last year, and this year its helped incrediblly. Its amazing. Thank you so so much.

[madness]

@Pi-thon
You are welcome.
Thanks for the kind words.

@Harry, check your email.

[Have done, thank you. I have replaced BOSS Technical Note 008 with the updated version, for which we are all truly thankful. All the best, Harry]

[momo]

now that the deadline has passed, what were the keys?

[kford_academy]

@momo, #92710

The keywords (in my experience) were NOTHING (keystream) and ZERO (square). This was done in the following way. ZERO was arranged in a square as shown:

_12345
1ZEROP
2QSTUV
3WXYAB
4CDFGH
5IKLMN

And each letter was substituted for a two-digit number in the form (row,column). (So Z=11,E=12,etc. I=J=51.)

Now NOTHING can be encrypted as:
55-14-23-45-51-55-44

This string of seven numbers was repeated to match the length of the ciphertext and then added to the relevant numbers in the encrypted plaintext. So the start of the message is encrypted:

Plaintext_ – _D _E _A _R __M _R …
Polybius__ – 42 12 34 13 _54 13 …
Keystream_ – 55 14 23 45 _51 55 …
Ciphertext – 97 26 57 58 105 68 …

[f6exb_the_frenchy]

ZERO and NOTHING

[id-hkj]

The polybius square used to encrypt it was as below:
1 2 3 4 5
1 Z E R O P
2 Q S T U V
3 W X Y A B
4 C D F G H
5 I K L M N
The keyword for that was ‘ZERO’

The thing that you had to subtract at the start was 55 14 23 45 51 55 44 – if you put that into the square, it gives you:
NOTHING

[The_Letter_Wriggler]

TYPE : Nihilist Cipher
GRID : 5 x 5
KEY1 : ZERO
KEY2 : NOTHING
MIXED: ZEROPQSTUVWXYABCDFGHIKLMN

\ 12345
1 ZEROP
2 QSTUV
3 WXYAB
4 CDFGH
5 IKLMN
———————————
Notes: mixed alpha variant used
ZEROPQSTUVWXYABCDFGHIKLMN instead of
ZEROABCDFGHIKLMNPQSTUVWXY

As if the cipher was not hard enough Harry used a variant mixed alphabet.
Well done Harry for the agony that caused!!!

[oliviacat3]

Now that the deadline has passed, did anyone manage to solve this one in excel? I tried initially but vlookup often gives errors when working with numbers and mid didn’t work to split the text as not all the numbers had the same number of digits.

[f6exb_the_frenchy]

Try with modulo 100 of numbers

[Jeffery_Bezos_the_second]

hello

[Author]

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