# How does the frequency work with Vigenère Cipher?

So I have an unknown key but with the length of 6. I understand that I calculate my own English letter frequency of my letters. I then compare it with the standard English letter frequency and identify a pattern shape similarity (in regards to a bar chart). I get my first letter for the key, however, how do I proceed with getting the second letter?

What do I change/do to get my second English letter frequency to compare to the standard English letter frequency?

thanks

• Did you calculate it for 6 sets? Commented Mar 2, 2021 at 23:47
• uh, sorry but what do you mean by calculate it for 6 sets? also now going over it, I think I've done it wrong. I basically used a website to calculate the amount of occurences a specific letter has, then I compared it to the standard english letter frequency for the first one. Commented Mar 3, 2021 at 0:05
• Since your Vigenère Cipher's key uses 6 different characters for the key, calculating the frequency of all letters are wrong. You need to find the frequency for 6 different sets. first sets positions 1,7,13,19,,,, second sets positions 2,8,14,20, etc. Commented Mar 3, 2021 at 0:13
• Ohh okay, I think I get it, gonna test it out now and see where I get, thank you! Commented Mar 3, 2021 at 0:50

So in your cipher text, the characters $$c_1, c_7, c_{13}, \ldots, c_{1+6i}, \ldots$$ should roughly obey the frequency statistics of a permuted or shifted standard English frequency. In the most standard Viginère, if you know E corresponds to cipher letter F (e.g. because it is (one of) the most frequent letters, the key letter is B and the alphabet has been shifted by $$1$$, allowing for immediate decryption of all letters in this subtext. The same holds for the characters for encrypted by second key letter $$c_2, c_8, c_{14}, \ldots c_{2+6i}, \ldots$$: check the histogram, to see if it looks like a shifted standard alphabet etc. In general, the cipher alphabets per key letter could be totally independent and then it can take quite a bit of hand work (guessing words and extrapolating from that) to completely solve the text, but if the standard alphabets are merely shifted, it's rather easy to determine the key letter.