National Cipher Challenge 2019 › Forums › National Cipher Challenge 2019: Countdown to Catastrophe › Welcome (back?) to the National Cipher Challenge
Tagged: Cracking the Vigenere cipher
30th September 2019 at 12:42 pm #40752
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.)
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Have fun!21st November 2019 at 4:41 pm #44282DavidsentMember
YO whats the answer to number 621st November 2019 at 4:41 pm #44289
That is indeed the question! …
Harry21st November 2019 at 5:17 pm #44291Cbrowngaldro-orgMember
Hello, could you give me some very brief guidlines to solve the vignere cypher21st November 2019 at 5:18 pm #4429621st November 2019 at 5:20 pm #44290DavidsentMember
pls21st November 2019 at 5:20 pm #44294Cbrowngaldro-orgMember
Yes………..22nd November 2019 at 10:31 am #44299OpdeadbushMember
Anyone got any hints to 6B? I’m stumped23rd November 2019 at 9:20 pm #44367MadnessMember
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:
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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