Vic Cipher Encoding and Decoding

So, here’s a quick rundown of how to use the Vic cipher with a given keyword, like “CIPHER”. We’ll need to create a matrix based on that keyword and then use it to encode and decode messages. Below, I’ll show you how to do that in Python!

def create_matrix(keyword):
    # Remove duplicates and create the base of the matrix
    keyword = ''.join(dict.fromkeys(keyword.upper()))
    alphabet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
    # Create the matrix with the keyword first
    matrix = list(keyword)
    for letter in alphabet:
        if letter not in matrix:
            matrix.append(letter)
    return [matrix[i:i + 6] for i in range(0, 36, 6)]  # Create 6x6 matrix

def encode(plaintext, keyword):
    matrix = create_matrix(keyword)
    encoded_message = ""
    
    for char in plaintext:
        if char.upper() in ''.join(matrix):
            index = ''.join(matrix).index(char.upper())
            row = index // 6
            col = index % 6
            encoded_message += f"{row}{col} "  # Using row and column as encoded form
        else:
            encoded_message += char  # Keep spaces and punctuation as is
    
    return encoded_message.strip()

def decode(encoded_message, keyword):
    matrix = create_matrix(keyword)
    decoded_message = ''
    pairs = encoded_message.split()
    
    for pair in pairs:
        if pair.isdigit() and len(pair) == 2:
            row, col = int(pair[0]), int(pair[1])
            decoded_message += matrix[row][col]
        else:
            decoded_message += pair  # Keep non-numeric characters as is
            
    return decoded_message

# Example usage
keyword = "CIPHER"
plaintext = "HELLO WORLD"
encoded = encode(plaintext, keyword)
decoded = decode(encoded, keyword)

print(f"Encoded: {encoded}")
print(f"Decoded: {decoded}")

Handling unexpected characters like spaces and punctuation means we just keep them in their original form when encoding and decoding. I used if char.upper() in ''.join(matrix) to check if the character exists in the matrix. If it does, we convert it; if not, we just append it as is.

As for performance, using a matrix of size 6×6 keeps things quick since you’re not dealing with massive data sets. The key is to efficiently find characters and handle them. If you ever run into memory or speed issues, consider tuning the algorithm or using optimized data structures.

Feel free to play around with this code and make it your own!