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Transformation

  • Description: I wonder what this really is...
  • Difficulty: Easy

🔎 Solution​

The challenge provides an encoded file and a short snippet of Python code:

''.join([chr((ord(flag[i]) << 8) + ord(flag[i + 1])) for i in range(0, len(flag), 2)])

This code is used to encode the original flag using the following logic:

  • The input string flag is split into pairs of characters (2 characters at a time).
  • For each pair flag[i] and flag[i+1], the code:
  • Converts both characters to their ASCII values using ord().
  • Shifts the first character's ASCII value 8 bits to the left (ord(flag[i]) << 8), which is equivalent to multiplying it by 256.
  • Adds the second character's ASCII value to it.
  • The sum is passed to chr(), converting it into a single Unicode character.
  • Finally, all Unicode characters are concatenated to form the encoded output.

Once the encoding logic is understood, decoding becomes a straightforward process - we simply reverse each step.

Each character in the encoded string represents a pair of original characters. To decode:

  • Convert the character back to its code point using ord().
  • Extract the high byte by shifting right 8 bits (>> 8).
  • Extract the low byte using a bitwise AND with 0xFF.
  • Convert both back to characters and concatenate.

Here's the Python script used for decryption:

def decrypt(encoded_str):
    decoded = ''
    for c in encoded_str:
        code_point = ord(c)
        high = code_point >> 8
        low = code_point & 0xFF
        decoded += chr(high) + chr(low)
    return decoded

with open('enc', 'r', encoding='utf-8') as f:
    encoded_data = f.read().strip()

decoded_flag = decrypt(encoded_data)
print(decoded_flag)

Running the script reveals the original flag.

🚩Flag​

picoCTF{16_bits_inst34d_of_8_e703b486}