Transformation picoCTF 2021 Solution

Description

1. Step 1Identify the encoding scheme
   Open enc in a text editor or print it - it contains Unicode characters with code points well above 127. This is a clue that multiple ASCII bytes were packed together. The encoding is: char = (ord(a) << 8) + ord(b), combining two consecutive ASCII bytes into one Unicode code point.
   Learn more

   Standard ASCII characters have code points from 0 to 127, fitting in 7 bits. A Unicode code point, however, can represent values up to 1,114,111. The encoding used here treats two ASCII bytes as the high byte and low byte of a 16-bit integer, producing a single Unicode character - effectively compressing two characters into one.

   This is not a standard encoding like UTF-16; it is a custom scheme. Looking at the code points of the characters in enc and noticing they are all in the range 0x2000 to 0x7e7e (printable ASCII pairs) is the main clue.

2. Step 2Reverse the encoding
   For each Unicode character c in enc, extract the high byte with (ord(c) >> 8) and the low byte with (ord(c) & 0xff). Convert each byte back to a character and concatenate - this reconstructs the original ASCII string containing the flag.
   python

   python3 -c "
   enc = open('enc').read().strip()
   print(''.join([chr(ord(c) >> 8) + chr(ord(c) & 0xff) for c in enc]))
   "

   Copied!
   Learn more

   The shift-right operator >> 8 drops the lower 8 bits, leaving the original high byte. The bitwise AND & 0xff masks to only the lower 8 bits, recovering the low byte. These are standard bit manipulation operations for unpacking multi-byte values - used extensively in binary format parsing and network protocol decoding.

   The challenge title "Transformation" refers to the encoding transformation applied to the plaintext. Reversing a transformation requires understanding the original operation - here a straightforward bit-packing scheme with no randomness or key.

Flag

The encoding pairs consecutive ASCII bytes into a single Unicode code point - reverse by extracting the high byte (>>8) and low byte (&0xff) separately.