Play Nice

Problem Description

Not all ancient ciphers were so bad... The flag is not in standard format. nc mercury.picoctf.net 30568 playfair.py

Points

110

Question Type

Cryptography

Hints

(None)

Approach

Understanding the Code

alphabet = open("key").read().rstrip()
m = generate_square(alphabet)
msg = open("msg").read().rstrip()
enc_msg = encrypt_string(msg, m)
print("Here is the alphabet: {}\nHere is the encrypted message: {}".format(alphabet, enc_msg))

This initializes important variables in the program. Running it in Terminal, we get that that the alphabet is "0fkdwu6rp8zvsnlj3iytxmeh72ca9bg5o41q" and enc_msg is "herfayo7oqxrz7jwxx15ie20p40u1i".

def generate_square(alphabet):
    assert len(alphabet) == pow(SQUARE_SIZE, 2)
    matrix = []
    for i, letter in enumerate(alphabet):
        if i % SQUARE_SIZE == 0:
            row = []
        row.append(letter)
        if i % SQUARE_SIZE == (SQUARE_SIZE - 1):
            matrix.append(row)
    return matrix

The cuntion generate_square creates a 2D list of letters and numbers, which is stored in the variable m. Printing it out, we get:

[['0', 'f', 'k', 'd', 'w', 'u'],
['6', 'r', 'p', '8', 'z', 'v'],
['s', 'n', 'l', 'j', '3', 'i'],
['y', 't', 'x', 'm', 'e', 'h'],
['7', '2', 'c', 'a', '9', 'b'],
['g', '5', 'o', '4', '1', 'q']]

def encrypt_string(s, matrix):
    result = ""
    if len(s) % 2 == 0:
            plain = s
    else:
            plain = s + "0fkdwu6rp8zvsnlj3iytxmeh72ca9bg5o41q"[0]
    for i in range(0, len(plain), 2):
            result += encrypt_pair(plain[i:i + 2], matrix)
    return result

The function encrypt_string simply adds a '0' to the end of the message if the length is odd and then runs encrypt_pair two characters in msg at a time. When solving, we can simply ignore this function and if the outputted message ends with a 0, test it with and without a '0' in the terminal.

def get_index(letter, matrix):
    for row in range(SQUARE_SIZE):
            for col in range(SQUARE_SIZE):
                    if matrix[row][col] == letter:
                        return (row, col)
    print("letter not found in matrix.")
    exit()

This function, get_index, gets the index of letter in matrix.

def encrypt_pair(pair, matrix):
    p1 = get_index(pair[0], matrix)
    p2 = get_index(pair[1], matrix)

    if p1[0] == p2[0]:
        return matrix[p1[0]][(p1[1] + 1)  % SQUARE_SIZE] + matrix[p2[0]][(p2[1] + 1)  % SQUARE_SIZE]
    elif p1[1] == p2[1]:
        return matrix[(p1[0] + 1)  % SQUARE_SIZE][p1[1]] + matrix[(p2[0] + 1)  % SQUARE_SIZE][p2[1]]
    else:
        return matrix[p1[0]][p2[1]] + matrix[p2[0]][p1[1]]

The function encrypt_pair is where characters in the message are changed. If the two letters are in the same row, both a shifted left, if they are in the same column, they are shifted down and if they are not in the same column or row, the column is switched.

Solving

I created a program playfair_solve to find the message

for i in range(0, len(enc_msg), 2):
    a = get_index(enc_msg[i], matrix)
    b = get_index(enc_msg[i+1],matrix)
    if(a[0] == b[0]):
        msg += matrix[a[0]][(a[1]-1) % SQUARE_SIZE] + matrix[b[0]][(b[1]-1)% SQUARE_SIZE]
    elif(a[1] == b[1]):
        msg += matrix[(a[0]-1)% SQUARE_SIZE][a[1]] + matrix[(b[0]-1)% SQUARE_SIZE][b[1]]
    else:
        msg += matrix[a[0]][b[1]] + matrix[b[0]][a[1]]

I looped through the letters in enc_msg two at a time and got their index in matrix, stored in a and b. If a[0] == b[0], then I would shift both of them left, if a[1] == b[1], I would shift them up and otherwise, I would switch b[1] and a[1]. The message that was outputted was "emf57mgc51tp693dtt4g3h7f8ouwq3". Running the program in terminal and inputting that value as plain text, it outputs "Congratulations! Here's the flag: 007d0a696aaad7fb5ec21c7698e4f754".

I tried picoCTF{007d0a696aaad7fb5ec21c7698e4f754} which didn't work and then tried 007d0a696aaad7fb5ec21c7698e4f754, which is the flag.

Flag

007d0a696aaad7fb5ec21c7698e4f754