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This topic contains 191 replies, has 48 voices, and was last updated by  Anonymous 2 years, 4 months ago.

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  • #34035

    Anonymous

    As we are very near the finish deadline, and a method of solution has been officially given by Joan, I hope this will get passed by the moderators(if not now then on Jan 4th!)

    There is no way we would ever have seen this as a Hill cipher, nor would we have seen the modular addition of adding the two letters together in the way described above. Our initial ideas were running key and auto key as natural development of the keyword for the Vigenere cipher. This was backed up by the large number of ciphertext letters for single letter words and the highly variable repeat distances between repeated threes, but we were snookered by the reversal of the cipher text for each word, so trying THE/AND/FOR etc for all the threes didn’t work at first – but once the reversal was tried then we found success with a trial plaintext FOR resulting in a giveaway GFO for the key – and we could start working backwards from there – the G implied an N, then this N implied an I and so on. We has a slight issue with every 12th letter (which scrambled our EXCEL implementation of autokey at first) that was relatively quickly resolved when we could deduce one or two of them by comparison with the plaintext. WE (perhaps naively) assumed that this was a self correction feature – one of the issues with auto key is that a single error garbles the rest of the message, so by ensuring every xth letter is cleartext, this would act as a reset point so that in the even of an error, the majority of the message could still be read. Once we added this to our EXCEL file the message resolved easily.

    A superb cipher to solve for this final challenge, and we will be looking forward to the 2018 edition 🙂

    #34044

    Anonymous

    <big>**** WARNING (TO HOPEFULLY PREVENT OTHERS LOSING THE PLOT) ****</big>

    After a lot of head scratching we had an epiphany this evening … the maths behind this particular “cipher” assumes that “A” is represented as 0 (zero), even though this has no Roman numeral equivalent. We’ve been using A=1, B=2, …, Z=26 all along and rolling the mod() results to suit.

    Adding in this context means (ASC(x)-65) + (ASC(y)-65), where ASC() returns the ASCII value for the given character (A=65, Z=90, a=97 and z=122).

    You can quickly generate the plain text using JavaScript, Python, Perl or even Excel / OpenOffice Calc.

    #34068

    Anonymous

    @Charles Airshine
    @theindianmeister_jh
    The plaintext is encoded as follows: for the third letter of ciphertext you add the second letter of plaintext to the third letter of plaintext, this gives the third letter of ciphertext. You need to work out how to decode with this info, but @Charles Airshine you should get real words (i.e. what you had done is incorrect).

    #34069

    Anonymous

    Thanks Joan – have got it now!

    #34071

    Anonymous

    The cipher was encrypted using the method described, you have to decrypt it, as it was encrypted by adding, you will have to subtract letters.

    #34072

    Anonymous

    Help I’m so annoyed I have 66%

    Joan would the matrices be:

    100000000000
    110000000000
    011000000000
    001100000000
    000110000000
    000011000000
    000001100000
    000000110000
    000000011000
    000000001100
    000000000110
    000000000011

    #34073

    Anonymous

    How do you reverse every word backwards?

    #34080

    Anonymous

    have you deciphered all of it? We did this in an earlier one and submitted a decrypt of about two thirds of the text having missed a page break out when copying and pasting!

    #34079

    Anonymous

    To explain to those struggling with the modular arithmetic (first, totally write a program to help you a numbers modulo 26. I did this and it’s the most useful thing in the world), I’ve written something with the cipher of 8b, and I’ll decode it so people understand the maths.

    The message is EBXMBAP.

    The trick for this is to number a as 0, b as 1, and z as 25, as opposed to a = 1, b = 2, z = 26. This is also how computers like it.

    I leave the first letter untouched, giving my current plaintext: e.
    I then convert B into a number modulo 26, which is 1, and my previous plaintext letter, e, into a number modulo 26, 4.
    I then perform 1 – 4 modulo 26, = 23 modulo 26. This corresponds to the letter x.
    My current plaintext is now: ex.

    I then convert X into a number modulo 26, which is 23, and my previous plaintext letter, x, into a number modulo 26, 23.
    I then perform 23 – 23 modulo 26, = 0 modulo 26. This corresponds to the letter a.
    My current plaintext is now: exa.

    I then convert M into a number modulo 26, which is 12, and my previous plaintext letter, a, into a number modulo 26, 0.
    I then perform 12 – 0 modulo 26, = 12 modulo 26. This corresponds to the letter m.
    My current plaintext is now: exam.

    I then convert B into a number modulo 26, which is 1, and my previous plaintext letter, m, into a number modulo 26, 12.
    I then perform 1 – 12 modulo 26, = 15 modulo 26. This corresponds to the letter p.
    My current plaintext is now: examp.

    I then convert A into a number modulo 26, which is 0, and my previous plaintext letter, p, into a number modulo 26, 15.
    I then perform 0 – 15 modulo 26, = 11 modulo 26. This corresponds to the letter l.
    My current plaintext is now: exampl.

    I then convert P into a number modulo 26, which is 15, and my previous plaintext letter, l, into a number modulo 26, 11.
    I then perform 15 – 11 modulo 26, = 4 modulo 26. This corresponds to the letter e.

    My plaintext is now: example.

    In practice this is actually much closer to an autokey/autoclave cipher, with a keyword of simply the letter A. Don’t forget that it stops every 12 characters!

    You could crack this by hand, like I did with this example, but it was 7 letters long and it still took me a while, and this ciphertext has a little under 4.5 thousand characters. So mechanise some part of the process with a program, even if it’s just a simple program which will do modulo 26 calculations for you. Stepping up from that, something that runs in 12s and stores each decrypted letter to decrypt the next ciphertext letter is great, and if you can mechanize the whole thing, all the better!

    #34083

    Anonymous

    once we find the secret message and we are on a team, do all members of the team have to send the secret message to the email or does only one have to? thanks

    #34084

    Anonymous

    @Emiloi88 asked “How do you reverse every word backwards?”

    It depends whether you are using pen & paper, spreadsheet, computer code etc.

    Pseudo-code to reverse words, but preserve word order could look something like:

    main(string){
    result=””;
    while (string.length>0){
    pointer=string.indexOf(” “);
    word=string.substring(0, pointer);
    string=string.substring(pointer);
    result+=reverse(word);
    }
    }

    reverse(word){
    result= “”;
    for(i = word.length-1; i >= 0; i–){
    result+=word[i];
    }
    return result;
    }

    #34087

    Anonymous

    Each team only needs to submit the secret message once.

    #34089

    Anonymous

    @TheCenturions remember that the Hill cipher may append letters on the end of sections so they can be completed, which requires removal afterwards.

    #34093

    Anonymous

    Help! We are back to school today and only have until 9:00 pm to do this because I was submitting answers for another challenge an 9:00 pm and it said that the deadline was reached and that it was 23:59. I don’t get the Hill cipher and my other team mates haven’t contacted me at all with 8B. Also, where do the pictures come in? They said Aquilae when I copied and pasted the pictures. Does that have anything to do with it.
    I am super stuck, one of my teammates may have solved it but for now it is unsolved.
    HELP!

    #34095

    Anonymous

    Joan are the matrices right?
    I posted them on this page

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