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9B HELP! please

Viewing 9 posts - 1 through 9 (of 9 total)
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  • #45136
    Helpme-1
    Participant

    I’ve split the ciphertext into 28 columns. Then I’ve added the columns which had 2s in it at various positions like 0,4,12 etc. Got a frequency analysis of 24 and then replaced the ciphertext with letters. But the index of coincidence is still not English like its more Vigenere like though I do not think its Vigenere. I’ve tried different reshuffling orders of the 7 columns which has 2s like 4,12,16 etc. Though I am able to now get 24 characters I am still not getting proper english like readable text. Not sure what I am missing. Please help need to solve this somehow before the answers are out. I Am even dreaming of numbers and columns. Its truly maddening now. 🙁 Thanks

    #45139
    Harry
    Keymaster

    Think Trifid (but adapted to use binary and ternary together as you have sort of figured out)

    Harry

    #45146

    I have organised it so one block of columns has 2s in it, and the other three don’t.
    Is the one with the twos in it encoded in a different way?

    #45145
    Macss
    Member

    I’m at the same exact stage as @Helpme-1. But I’m still confused on this ternary and binary together thing. Are we supposed to use quadgrams (I think that’s the right word) going horizontally written in ternary, because that’s what I’m thinking about. If I manage to figure this ternary-binary thing out I can finally solve it as a Vigenere.

    #45143

    So you figured out that 7 columns of each group of 28 contain a 2. What now? For 24 characters, you need 0/1, 0/1, 0/1/2 and 0/1. Each letter has to contain at most one 2 and always in the same position. Now you need to experiment with different ways to take 4 such characters. Imagine if this below was a sample of the cipher (it’s not):

    0 1 1 0 1 0 1 1 0 1 1 0 1 1 0 2 1 2 2 0 2 0 1 0 0 0 1 0
    1 0 1 1 0 0 1 0 0 0 0 1 0 0 2 0 0 0 1 1 1 0 0 0 1 0 1 1

    You could take positions 0, 7, 14, 21; 1, 8, 15, 22 etc from the first row and then move on to the next. That would be 0100, 1021, 1110…
    Or, you could take positions 0, 7, 14, 21 from all rows and then add 1 to the positions. That would yield 0100, 1020…

    Once you have the 4 digit combinations, assign a random letter to each unique one. Check the letter frequencies and IoC to see whether they match English. Then, monoalphabetic substitution!

    Good luck.

    #45142
    Twisha-k
    Member

    Thanks for the hint Harry! I finally got it…it was sooo interesting and I really enjoyed it now that I got it😂looking forward to next year!

    #45148
    Harry
    Keymaster

    Glad to be of service!

    All the best,

    Harry

    #45157
    Leo_yates
    Member

    i Have the cipher text out into 28 columns as can be seen here https://reddamhouse-my.sharepoint.com/:x:/g/personal/leo_yates_reddamhouse_org_uk/EU4nu2ljPD5GrX-zfUoZYKkBIyc276nHmxeXH76C5HPPPQ?e=39pFgC
    but i am not getting 7 zeros in each row as stated by other members of the challenge i have been working on theis all winter and i really would like to crack it before the end.
    thanks

    #45155

    I have found a combination that has 24 combinations and english like frequencies with my little python program, but i am baffled by the amount of different ways you could get these 4 digit numbers
    After i substituted what I had, I wasent able to make it make sense in English, (though it did look like it was trying to make sense-probaly just me)

    Also, just to check I am on the right track, should there be the same amount of characters in the final text as lines in the 28 thing (660)

    I really want to get it finished for the final day,
    Thanks

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