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Welcome (back?) to the National Cipher Challenge

National Cipher Challenge 2019 Forums National Cipher Challenge 2019: Countdown to Catastrophe Welcome (back?) to the National Cipher Challenge

Viewing 9 posts - 1 through 9 (of 9 total)
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  • #40752
    Harry
    Keymaster

    Its good to have you back if we have met you before, and great to welcome you if you haven’t. The forum is a place to hang out and discuss anything related to the Challenge and quite a few things that are not. (It is a sort of virtual version of the Mathematical Sciences Student Centre that we have here at Southampton.)

    There are not many house rules.
    Be kind – people on the web are still people and have feelings.
    Don’t write anything you wouldn’t be proud to be associated with. Most people don’t know who you are, but some do!
    Don’t publish solutions or hints without clearing them with Harry first! It ruins the competition for everyone. (And don’t think you will get away with publishing them somewhere else on the web either. Harry can use Google too!)
    Do tell us your best maths jokes. There aren’t many good ones, but you will get a very receptive audience here.
    Also challenge us with your best maths puzzles, but don’t try to publish encrypted messages as Harry and the Elves will assume they contain something scurrilous and block them!
    Be polite to the Elves. They have a lot to do and are trying their best to support you by running the forum.
    Have fun!

    Harry's signature

    #44282
    Davidsent
    Member

    YO whats the answer to number 6

    #44289
    Harry
    Keymaster

    That is indeed the question! …

    Harry

    #44291

    Hello, could you give me some very brief guidlines to solve the vignere cypher

    #44296
    Harry
    Keymaster

    How about looking at the Forum post:

    #44290
    Davidsent
    Member

    pls

    #44294

    Yes………..

    #44299
    Opdeadbush
    Member

    Anyone got any hints to 6B? I’m stumped

    #44367
    Madness
    Member

    @Cbrowngaldro-org

    If you don’t want to read the PDF, or if you don’t want to use a computer, then the book by
    Gaines is a good resource for breaking Vigenere ciphers by hand.

    The first thing you have to do is figure out the period, which is the same as the key length.
    Gaines says to look for repeating groups of characters; the distance between them (from the
    start of one to the start of the other) should be a multiple of the period.

    But I just thought of another way, if you like statistics. Step 1: guess at the period. Step 2:
    count each letter in a slice using that period; so, for example, if you guessed 6, then take
    every sixth letter from the ciphertext and tally the As, the Bs, … and the Zs. Step3: find the
    least frequent letter in your tallies. It should have a zero tally, or at most 1% of the total.
    If not, you probably have the wrong period and have to guess again. If you like math, then you
    might also realize that factors are important. So if 6 is wrong, then so are 2 and 3. But if
    6 passes the test, then the best answer might be 2 or 3, so you should check them, too.

    Once you have the period, let’s say it’s 5, you need to make a tally for each slice. So the
    first slice is every 5th letter starting with the first letter of the ciphertext. Make a table
    of how often each letter occurs in it. The second slice is every 5th letter, starting with the
    second letter of the ciphertext. Make a table for it. Do as many tables as the period/keylength.

    For each slice, shift your table against a (doubled) table of standard letter frequencies until
    you get a close match. You can find a graph of the standard freqencies for English here:
    http://practicalcryptography.com/ciphers/caesar-cipher/
    Here is my attempt to make a graph to exemplify the process, but it might not line up right,
    depending on the font.

    5 5 0 1 2 1 5 5 1 0 5 5 5 2 1 2 0 1 0 5 1 1 3 9 1 1
    your table: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
    standard: a b c d e f g h i j k l m n o p q r s t u v w x y z a b c d e f g h i j k l …
    6 1 1 2 9 1 1 5 5 0 0 2 1 5 5 1 0 5 5 6 2 1 2 0 1 0 6 1 1 2 9 1 1 5 5 0 0 2 …

    When they line up, then look at where the ‘a’ is in the standard list. The letter in your table
    next to it is the key for that slice. In my example, it is ‘T’.

    When you have all the keys for the slices, you have the key to the whole cipher. Then
    decrypting is easy, but tedious.

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