C4ptur3 Th3 Fl4g // #bobbybrown

This is a beginner-level CTF created by dcdavidlee.

There is no virtual machine deployment necessary.

TO DO : I am yet to complete part 10 of Task 1. But all the other tasks are completed.

Task 1 - Translation and Shifting

This task requires the user to adopt a number of different strategies to find the hidden gems - translate, shift and/or decode.

Part 1 - Substitution Cipher

c4n y0u c4p7u23 7h3 f149?

This is a basic substitition cyber:

Replace ‘4’ with ‘a’
Replace ‘0’ with ‘o’
Replace ‘7’ with ’t'
Replace ‘2’ with ‘r’
Replace ‘3’ with ‘e’
Replace ‘9’ with ‘g’

dict = {
    '0' : 'o',
    '2' : 'r',
    '3' : 'e',
    '4' : 'a',
    '7' : 't',
    '9' : 'g',
}

encoded = 'c4n y0u c4p7u23 7h3 f149?'
decoded = [dict.get(x,x,) for x in encoded]

print('Decoded string : '.format(''.join(decoded)))

Part 2 - ASCII encoder

01101100 01100101 01110100 01110011 00100000 01110100 01110010 01111001 00100000 01110011 01101111 01101101 01100101 00100000 01100010 01101001 01101110 01100001 01110010 01111001 00100000 01101111 01110101 01110100 00100001

Converting this binary code to decimal and revealing it with an ASCII table will reveal its secrets. For an explanation on how to convert from binary to decimal, you can read this article.

from strings import ascii_letters

def convert(binary):
    sum = 0
    for k,v in enumerate(binary) :
        sum += (int(v) * pow(2, 7-k))
    return sum

if __name__ == "__main__":
    code = '''
        01101100 01100101 01110100 01110011 00100000 
        01110100 01110010 01111001 00100000 01110011 
        01101111 01101101 01100101 00100000 01100010 
        01101001 01101110 01100001 01110010 01111001 
        00100000 01101111 01110101 01110100 00100001'''

    decoded = ''.join(
        [ chr(convert(x)) for x in code.split() ])

    print(decoded)

Part 3 - Base32 encoding

MJQXGZJTGIQGS4ZAON2XAZLSEBRW63LNN5XCA2LOEBBVIRRHOM======

This string is in base32 encoding. To learn more about base encoding, you can read this fantastic article.

A quick decoding script in python will reveal the hidden secret.

from base64 import b32decode

encoding = 'MJQXGZJTGIQGS4ZAON2XAZLSEBRW63LNN5XCA2LOEBBVIRRHOM======'
print(b32decode(encoding).decode('UTF-8'))

Part 4 - Base64 encoding

RWFjaCBCYXNlNjQgZGlnaXQgcmVwcmVzZW50cyBleGFjdGx5IDYgYml0cyBvZiBkYXRhLg==

Similarly, this string is a base64 encoding. A quick decoding script in python will reveal the hidden secret.

from base64 import b64decode

encoding = 'RWFjaCBCYXNlNjQgZGlnaXQgcmVwcmVzZW50cyBleGFjdGx5IDYgYml0cyBvZiBkYXRhLg=='
print(b64decode(encoding).decode('UTF-8'))

Part 5 - Base16 encoding

68 65 78 61 64 65 63 69 6d 61 6c 20 6f 72 20 62 61 73 65 31 36 3f

This is base16 (hexadecimal) encoding. These are paired as bytes to make it easier to read.

We can use python’s bytes class to convert from hex to the hidden message.

encoding = '68 65 78 61 64 65 63 69 6d 61 6c 20 6f 72 20 62 61 73 65 31 36 3f'
encoding = ''.join(encoding) # Assembling it as one long string

print(
    bytes
    .fromhex(encoding)
    .decode('UTF-8'))

Part 6 - ROT 13

Ebgngr zr 13 cynprf!

The above is a ROT-13 cipher which rotates each alpha character 13 times around the standard 26 alphabet. For non-alpha characters, that is, characters that are either special symbol or numeric, they are left as is.

from string import ascii_lowercase

encoded = "Ebgngr zr 13 cynprf!".lower() # Putting them all to lowercase makes it easier to manage.

# Remove the value of 'a' from each alpha character value to map it to the array of lowercase letters, 
# then move each new value 13 characters forward to base 26 (26 letters in the alphabet).
# Returns an array of positional elements in the lowercase letter array. Characters that are not alpha
# (i.e. numbers or special characters) are left unaffected.

decoded_indices = [(ord(l) - ord('a') + 13) % 26 if l.isalpha() else l for l in encoded]

# Map array of positions (decoded_numeric) to the lowercase letters if they are digits else,
# return the character as-is.
# Join the elements in the array to create a single string statement.

decoded = ''.join(
    [ 
        ascii_lowercase[l] if type(l) is int else l for l in decoded_indices
    ]
)

print(decoded)

Part 7 - ROT-47

*@F DA:? >6 C:89E C@F?5 323J C:89E C@F?5 Wcf E:>6DX

Similarly to ROT-13, ROT-47 uses the ASCII chart but only a 94-character subset of alphabets between the character 33 (!) and the character 126 (~).

You can read about the ROT-47 and its pseudocode by going to this link.


encoded = '*@F DA:? >6 C:89E C@F?5 323J C:89E C@F?5 Wcf E:>6DX'
decoded = []

for character in encoded:
    char_code = ord(character)
    if(char_code > 32 and char_code <= 80): # Do not include 32 in the range to ignore spaces
        decoded.append(chr(char_code + 47))
    elif(char_code >= 79 and char_code <= 127):
        decoded.append(chr(char_code - 47))
    else:
        decoded.append(character)

print(decoded)

Part 8 - Morse Code

- . .-.. . -.-. — – – ..- -. .. -.-. .- - .. — -.

. -. -.-. — -.. .. -. –.

The above are two messages in Morse Code. Having a “Morse Code” dictionary would help in translating these encodings back into something we can read.

morse_dict = {
    '.-': 'A', '-..': 'D', '-.-.': 'C', '.': 'E', '..-.': 'F', 
    '--.': 'G', '....': 'H', '..': 'I', '.---': 'J', '-.-': 'K', 
    '.-..': 'L', '--': 'M', '-.': 'N', '---': 'O', '.--.': 'P', 
    '--.-': 'Q', '.-.': 'R', '...': 'S', '-': 'T', '..-': 'U', 
    '...-': 'V', '.--': 'W', '-..-': 'X', '-.--': 'Y', '--..': 'Z', 
    '.----': 1, '..---': 2, '...--': 3, '....-': 4, '.....': 5, 
    '-....': 6, '--...': 7, '---..': 8, '----.': 9, '-----': 0
}

morse_codes = list([
    '- . .-.. . -.-. --- -- -- ..- -. .. -.-. .- - .. --- -.', 
    '. -. -.-. --- -.. .. -. --.'
])

for ms in morse_codes:
    print(''.join([morse_dict.get(x, '') for x in ms.split()]))

Part 9 - ASCII encoding

85 110 112 97 99 107 32 116 104 105 115 32 66 67 68

This is basic ASCII encoding. Compare each number to the ASCII table to obtain the flag.

encoded = "85 110 112 97 99 107 32 116 104 105 115 32 66 67 68".split()
print(''.join([chr(int(x)) for x in encoded]))

Part 10 - Multiple Ciphers

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This is a multi-cipher problem and requires one to use all the learnings from the previous tasks:

Cipher 1: Base64
Cipher 2: Morse code
Cipher 3: Binary to ASCII
Cipher 4: ROT 47
Cipher 5: BCD to ASCII


[To do]

Task 2 - Spectogram

Wikipedia defines a spectogram as “a visual representation of the spectrum of frequencies of a signal as it varies with time. When applied to an audio signal, spectrograms are sometimes called sonographs, voiceprints, or voicegrams.”.

We can therefore visualize this file using any wave analyzing progam, such as Audacity. Changing waveform view to Spectrogram under the name drop down reveals the secret message:

spek_audacity

Task 3 - Steganography

Steganography is the practice of concealing a file, message, image, or video within another file, message, image, or video. Steganography looks to conceal secrets in plainsight. It therefore isn’t the same as cryptography that applies mathematical techniques to ensure privacy.

For this task, using a steganography application like steghide will help reveal the hidden file. In this simple instance, a password was not needed to extract the hidden file from the steg file.

steghide extract -sf stegosteg.jpg

Task 4 - Security Through Obscurity

Similarly to Task 3, the challenges uses steganography to hide files. But, because the passphrase is not obvious, we cannot use steghide directly to extract the file.

Instead, we can try and read out the contents of the file to see whether there are any familiar strings that can help us identify the hidden file. Using the strings command, we are able to see some information towards the end of the file.

strings meme.jpg

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