Untitled :: The Examples Book

Introduction

You got a letter from the agency, and it’s encrypted to protect the information from prying eyes.

We need to decrypt the letter to read the content.

What is Vigenère Cipher?

It’s a method of encryption… more details to come. For now, this article is a good place to start.

Click to see the full Vigenère Cipher table

Every letter in the alphabet has its numeric value, where a being 0 and increment of 1.

Practice Exercises

Encryption

The text sample is "Billy loves riding his bike."
The keyword is "YILQ".

To make the encryption easier, we’ll remove punctations and do uniform capitalization (either uppercase or lowercase all the words). Then, we will set up our new text using the repetation of the keyword.

BILLY LOVES RIDING HIS BIKE
YILQY ILQYI LQYILQ YIL QYIL

Now, we’ll encode using both sentences and the Vigenère Cipher table.

How do we look up the table?
* Horizontal: text sample letters
* Vertical: keyword letters
* Intersection: encrypted letters

The encrypted letter for our first letter is Z.

Click to see how we look up for the first letter using the Vigenère Cipher table

The encrypted letter for our second letter is Q.

Click to see how we look up for the second letter using the Vigenère Cipher table

After getting all the letters from the table, the new (encrypted) line should be

Click to reveal the answer

ZQWBW TZLCA CYBQYW FQD RGSP

Decryption

The encrypted text sample is "WHRTI DRR FMIFRTIQT FKDHS BH MQFVPMWY."
The keyword is "DANCE."

How do we look up the table?
* Horizontal: text sample letters
* Vertical: keyword letters
* Intersection: encrypted letters

To find the original letter, start with the letter from the keyword (horizontal) and look for the encrypted letter on that row. The column of the encrypted letter is the original letter. It’s the same process as encryption except it’s backwards.

WHRTI DRR FMIFRTIQT FKDHS BH MQFVPMWY
DANCE DAN CEDANCEDA NCEDA NC EDANCEDA

The first letter of the original unencrypted text is T.

Click to see how we look up for the first letter using the Vigenère Cipher table

The fifth letter of the original unencrypted text is E.

Click to see how we look up for the first letter using the Vigenère Cipher table

After getting all the letters from the table, the new (decrypted) line should be

Click to reveal the answer

THERE ARE DIFFERENT SIZES OF INFINITY

Here’s the letter.

TOP SECRET
CLASSIFIED INFORMATION for authorized personnel only

EB YEK XMVR TNWLKZP BF QQ WBKIFPQFR LDIK E TKLP ASO NFYFZ.

Q RQ SOAZKFEVX CGQ IEH QKCI XWWU KS LDQJ TGOAZFDU ULVVAZ TEKA.

TFGSPQFR AO BYMJPG VMYDB GSAJB VMYDB VMYDB EMFA NFYJ OMMIF AQXLL WVU RWCIKMNA AVZWJBP WWRME TGEVK DWNW KLJAM JMP KVV SFA WEI.

YKWU PMYS.

keyword: WIRES

Is there a shortcut?

If you did the practice exercises, you probably have a better understanding how Vigenère Cipher works, but it’s very time-consuming to encrypt or decrypt one sentence. I can’t imagine how long it will take us to decrypt the whole text.

Lucky for us, there is a shortcut. There is a formula that describes the relationship between the encrypted letter, the keyword, and the original letter.

Equations:
Encryption = (Original + Keyword) % 26
Decryption = (Encryption - Keyword + 26) % 26

The percentage (%) represents the modulo function - calculates the remainder after the division of two numbers.

Let’s decrypt the letter in Python.

We are ignoring the first two lines since they are already readable ('decrypted').

# Remove punctations and transform the text into a long sentence
text = "EB YEK XMVR TNWLKZP BF QQ WBKIFPQFR LDIK E TKLP ASO NFYFZ Q RQ SOAZKFEVX CGQ IEH QKCI XWWU KS LDQJ TGOAZFDU ULVVAZ TEKA TFGSPQFR AO BYMJPG VMYDB GSAJB VMYDB VMYDB EMFA NFYJ OMMIF AQXLL WVU RWCIKMNA AVZWJBP WWRME TGEVK DWNW KLJAM JMP KVV SFA WEI YKWU PMYS"
# our keyword is WIRES
keyword = "WIRES"
# initialize an empty string for our decrypted string
decrypted = ""

We can create a dictionary in Python where every letter has a numeric representation.

#key:value

# Set up a dictionary in Python
# The format is key:value (just like how you'd look up a word, word:meaning)
# Suppose I look up for the letter, 'o', I'll get the number, 14,
alphabet = {"A":0,"B":1,"C":2,"D":3,"E":4,"F":5,"G":6,"H":7,"I":8,"J":9,
            "K":10,"L":11,"M":12,"N":13,"O":14,"P":15,"Q":16,"R":17,"S":18,"T":19,
            "U":20,"V":21,"W":22,"X":23,"Y":24,"Z":25}

We can convert the mathematical equation for decryption into Python code below.

# pointer for the keyword since it'll be used repeatly
position = 0;

# for loop to look at every letter in the sentence, text
for letter in text:
    # check to make sure the letter is not a space character (if it is, jump to else condition)
    if letter != ' ':
        # check that the pointer doesn't go past the keyword length
        if position >= len(keyword):
            # otherwise, reset the pointer back to the beginning of the keyword (e.g., 0)
            position = 0;
        # calculate the position where the decrypted letter is in the alphabet
        decrypted_number = (alphabet[letter] - alphabet[keyword[position]] + 26) % 26
        # search through the dictionary to find the letter
        for key,value in alphabet.items():
            # if value in dictionary matches the decrypted number you calculated
            if value == decrypted_number:
                # add the letter to your decrypted string
                decrypted += key
        # update the position to move to next letter in the keyword
        position += 1
    # otherwise, just add the space character directly to the encrypted sentence
    else:
        decrypted += ' '

Here’s the code snippet for encryption (very similiar to the snippet above).

# pointer for the keyword since it'll be used repeatly
position = 0;

# for loop to look at every letter in the sentence, text
for letter in text:
    # check to make sure the letter is not a space character (if it is, jump to else condition)
    if letter != ' ':
        # check that the pointer doesn't go past the keyword length
        if position >= len(keyword):
            # otherwise, reset the pointer back to the beginning of the keyword (e.g., 0)
            position = 0;
        # calculate the position where the decrypted letter is in the alphabet
        encrypted_number = (alphabet[letter]+alphabet[keyword[position]]) % 26
        # search through the dictionary to find the letter
        for key,value in alphabet.items():
            # if value in dictionary matches the encrypted number you calculated
            if value == encrypted_number:
                # add the letter to your decrypted string
                encrypted += key
        # update the position to move to next letter in the keyword
        position += 1
    # otherwise, just add the space character directly to the encrypted sentence
    else:
        encrypted += ' '

Try to run the encryption code snippet with those variables.

# Remove punctations and transform the text into a long sentence
text = "STOP WAITING FOR EXCEPTIONAL THINGS TO JUST HAPPEN"
# our keyword is PINK
keyword = "PINK"
# initialize an empty string for our encrypted string
encrypted = ""

Click to check your work for encryption

After running the code, your encrypted variable should be

HBBZ LIVDXVT PDZ RHRMCDXWAKA BUSCOF DD RHCI PNZEMA