# Coded Correspondence help

I am currently working on the project Coded Correspondence.. I was doing well until I got to step 5. Here are the instructions.

Great work! While you were working on the brute force cracking of the cipher, Vishal sent over another letter. That guy is a letter machine!

Salutations! As you can see, technology has made brute forcing simple ciphers like the Caesar Cipher extremely easy, and us crypto-enthusiasts have had to get more creative and use more complicated ciphers. This next cipher I’m going to teach you is the Vigenère Cipher, invented by an Italian cryptologist named Giovan Battista Bellaso (cool name eh?) in the 16th century, but named after another cryptologist from the 16th century, Blaise de Vigenère.

The Vigenère Cipher is a polyalphabetic substitution cipher, as opposed to the Caesar Cipher which was a monoalphabetic substitution cipher. What this means is that opposed to having a single shift that is applied to every letter, the Vigenère Cipher has a different shift for each individual letter. The value of the shift for each letter is determined by a given keyword.

Consider the message:

``````barry is the spy
``````

If we want to code this message, first we choose a keyword. For this example, we’ll use the keyword

``````dog
``````

Now we repeat the keyword over and over to generate a keyword phrase that is the same length as the message we want to code. So if we want to code the message “barry is the spy” our keyword phrase is “dogdo gd ogd ogd”. Now we are ready to start coding our message. We shift each letter of our message by the place value of the corresponding letter in the keyword phrase, assuming that “a” has a place value of 0, “b” has a place value of 1, and so forth.

``````          message:    b  a  r  r  y    i  s    t  h  e    s  p  y

keyword phrase:    d  o  g  d  o    g  d    o  g  d    o  g  d
``````

resulting place value: 24 12 11 14 10 2 15 5 1 1 4 9 21

So we shift “b”, which has an index of 1, by the index of “d”, which is 3. This gives us an place value of 24, which is “y”. Remember to loop back around when we reach either end of the alphabet! Then continue the trend: we shift “a” by the place value of “o”, 14, and get “m”, we shift “r” by the place value of “g”, 15, and get “l”, shift the next “r” by 4 places and get “o”, and so forth. Once we complete all the shifts we end up with our coded message:

``````ymlok cp fbb ejv
``````

As you can imagine, this is a lot harder to crack without knowing the keyword! So now comes the hard part. I’ll give you a message and the keyword, and you’ll see if you can figure out how to crack it! Ready? Okay here’s my message:

``````txm srom vkda gl lzlgzr qpdb? fepb ejac! ubr imn tapludwy mhfbz cza ruxzal wg zztcgcexxch!
``````

and the keyword to decode my message is

``````friends
``````

Because that’s what we are! Good luck friend!
And there it is. Vishal has given you quite the assignment this time! Try to decode his message. It may be helpful to create a function that takes two parameters — the coded message and the keyword — then work towards a solution from there.

Below is my code for the function. The problem is when it makes the keyword phrase it doesn’t skip the spaces and I’m not sure how to do that. Any help would be appreciated. See my code below.

def decoding_machine_1(message, keyword): alphabet = 'abcdefghijklmnopqrstuvwxyz' punctuation = " .,?!'" keyword_message = "" coded_message = "" for i in range(len(message)): if message[i] in alphabet: keyword_message += keyword[i % len(keyword)] elif message[i] in punctuation: keyword_message += test_message[i] for i in range(len(test_message)): if test_message[i] in alphabet: letter_index = (alphabet.index(test_message[i])-alphabet.index(keyword_message[i])) % 26 print(letter_index, test_message[i],keyword_message[i]) coded_message += alphabet[letter_index] elif test_message[i] in punctuation: coded_message += test_message[i] return print(coded_message)

You can use a seperate counter to iterate through the keyword (instead of using `i`)

``````  test_message = message
j = 0  # used to iterate through the keyword
for i in range(len(message)):
if message[i] in alphabet:
keyword_message += keyword[j % len(keyword)]
j += 1  # go to next letter in keyword if used a letter
elif message[i] in punctuation:
keyword_message += test_message[i]
``````

demonstration.

def decoding_machine_1(message, keyword): alphabet = 'abcdefghijklmnopqrstuvwxyz' punctuation = " .,?!'" keyword_message = "" coded_message = "" test_message = message j = 0 for i in range(len(message)): if message[i] in alphabet: keyword_message += keyword[j % len(keyword)] j += 1 elif message[i] in punctuation: keyword_message += test_message[i] print(keyword_message) for i in range(len(test_message)): if test_message[i] in alphabet: letter_index = (alphabet.index(test_message[i])-alphabet.index(keyword_message[i])) % 26 print(letter_index, test_message[i],keyword_message[i]) coded_message += alphabet[letter_index] elif test_message[i] in punctuation: coded_message += test_message[i] return coded_message message = "barry is the spy" keyword = "dog" print(decoding_machine_1(message, keyword))

Also, at the end,
`return print(coded_message)`
should be
`return coded_message`

And why do you have a separate `test_message` variable in the function?
You can call the function with any message string as an argument.