8b
Tagged: 8b
- This topic has 17 replies, 12 voices, and was last updated 5 months, 3 weeks ago by The-Letter-Wriggler.
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1st December 2022 at 10:53 pm #87418someoneParticipant
any help on 8b
3rd December 2022 at 9:01 am #87422YeetcadamyParticipantI think it’s a form of a transposition cipher, because the frequency of letters matches the english language, but I cannot get it to work. The size of groups is either 5 or 7 as the total number of characters is 5 or 7
5th December 2022 at 12:06 pm #87425Ellie05Participanti would my friend but im just as stuck as you 🙁
5th December 2022 at 12:07 pm #87434A1PHA_L3G3NDParticipantFrom the helpful comment given by Madness, it could be some version of the porta cipher called the quagmire IV to be specific.
[Just to be clear, posting this does not mean it is right! Harry
ps (or wrong!)]5th December 2022 at 12:07 pm #87437sbParticipantThis one is really challenging. I do think it’s a transposition but I can’t make it work. I’ve got the first 5 letters – I think – but the rest of it doesn’t make sense. I could really appreciate some help with it.
5th December 2022 at 12:07 pm #87436f6exbParticipantThe size of groups can also be 35 or 41, but I cannot get it to work.
5th December 2022 at 1:24 pm #87444f6exbParticipantAfter debugging my program, so far so good.
5th December 2022 at 1:25 pm #87445The-Letter-WrigglerParticipantThis is embarrasing, I too have not yet solved 8B.
The tips say: Factors are your friend!
I broke the text into blocks and encrypted each block separately. . .The factors are 5, 7 and 41.
I have tried both a Normal Transposition and
a Columnar Transposition in the following ways
Write by Rows Read by Columns (normal way)
Write by Rows Read by Rows
Write by Columns Read by Rows
Write by Columns Read by ColumnsUsing 35 blocks of 41 letters
I have only been able to obtain the first 8 WORDSIs this a new take on the transposition cipher Harry?
[Beginning to look like it! Harry]
5th December 2022 at 3:46 pm #87449madnessParticipant1. It is not a Porta or even a quagmire.
2. It is a transposition.
3. Each block is encrypted the same way.
4. The key is short enough that you can brute-force it.
5. The blocks are 35 letters.Harry can censor as much of that as he likes.
5th December 2022 at 5:24 pm #87451f6exbParticipant@TLW
“Using 35 blocks of 41 letters
I have only been able to obtain the first 8 WORDS”How can you write 41 letters in a rectangle ?
6th December 2022 at 8:54 am #87454ciphercutterParticipantCensor as required.
I always start transposition ciphers by breaking into a grid.
Am I aiming for 35 across the top by 41 down?
41 across and 35 down?
Or something else entirely?6th December 2022 at 8:54 am #87455The-Letter-WrigglerParticipantFunny old world isn’t it – five mins after my post above I solve it, duh!
6th December 2022 at 9:00 am #87457msjjmathsicloud-com-1ParticipantEureka. Thanks for the pointers Madness. Made some progress at the weekend but your post today helped me immensely.
6th December 2022 at 9:01 am #87456ADecipherParticipantWhat are blocks in a transposition cipher?
[The blocks are the units you break the letters into in order to encrypt it.
So if the text had sixty letters, you might break it into 4 blocks of fifteen letters each, then encrypt each block separately:
THEBL
OCKSA
RETHEUNITS
YOUBR
EAKTHELETT
ERSIN
TOINORDERT
OENCR
YPTITHope this helps, Harry]
6th December 2022 at 9:01 am #87458ciphercutterParticipantWell feel free to censor the entirety of my previous transmission. I have finally cracked it.
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