HIB CHB Finals 2026 - Forensics & Reverse Engineering Writeup

1. Challenge Overview

Author: FAZer0
Event: Hack It Brawijaya Finals 2026
Date: 16 Agustus 2026

2. Forensics: 私たちの記憶の停止 (752 pts)

Investigation

Analyzing the Android artifact dump reveals an encrypted document Lyra_Poros_CV.docx.Hack-It-Braw. The Termux .bash_history contains the encryption script:

KEY=$(echo -n "https://open.spotify.com/playlist/7F5tC5VYL5Pj075kBaQM1b?si=r84XWSPMRk-T0A5-MHXVoA" | sha256sum | cut -d' ' -f1)
IV=$(echo -n "Is it painful to be the person who waits? Or is it more painful to be the person who makes others wait? Either way, there's no need to wait anymore. That's what is most painful." | sha256sum | cut -d' ' -f1 | head -c 32)
SALT="KIIRO"

encrypt_file() {
    local file="$1"
    local out_file="$file.Hack-It-Braw"
    {
        echo "${SALT}"
        openssl enc -aes-256-cbc -K "$KEY" -iv "$IV" -nosalt -in "$file" 2>/dev/null
    } > "$out_file"
    rm -f "$file"
}

Decryption Parameters

• KEY (Hex): 60319782eba030d4a2f7f4fcf60e664e68bfa41f4ed2d22655e0512b76f79356
• IV (Hex): e14932e40402d141698e98c52ded6374
• Pre-pended Salt: Strip first 6 bytes (KIIRO\n) from the ciphertext before performing AES-256-CBC decryption in CyberChef.

3. Reverse: Marshmallow (1,000 pts)

Investigation & Header Repair

Given marshmallow.pyc with missing magic header bytes. Inspecting with xxd shows zeroed header offsets.

1. Adding Python 3.11 standard magic number header a7 0d 0d 0a 00 00 00 00 00 00 00 00 00 00 00 00.
2. Decompiling via PyLingual recovers the core validation routine:

import sys

KEY = b't04st3d'
TARGET = b'\xe1\xd2\xc1\x96\xfd\xae\xdcC\x92!HO\x0e\x15\x92K\xa8w\xb6\xe8,\xfb4\xfc\xa8\xaf\xc6\xfd\x84F\xd4\xf3\xe2\xe9n-w\x05\xe2K{/\xe6M\xce\xab'

def toast(candidate):
    out = []
    for i, ch in enumerate(candidate):
        x = ord(ch)
        x ^= KEY[i % len(KEY)]
        x = ((x << 3) | (x >> 5)) & 255
        x = (x + i * 7) & 255
        out.append(x)
    return bytes(out)

Solver / Inversion Script

Since each character is transformed independently without cross-character state, we can directly invert the bit operations or brute-force character-by-character:

KEY = b't04st3d'
TARGET = b'\xe1\xd2\xc1\x96\xfd\xae\xdcC\x92!HO\x0e\x15\x92K\xa8w\xb6\xe8,\xfb4\xfc\xa8\xaf\xc6\xfd\x84F\xd4\xf3\xe2\xe9n-w\x05\xe2K{/\xe6M\xce\xab'

flag = []
for i, target_byte in enumerate(TARGET):
    # Invert (x + i*7) & 255
    x = (target_byte - i * 7) & 255
    # Invert bit rotate left 3 -> rotate right 3
    x = ((x >> 3) | (x << 5)) & 255
    # Invert XOR
    char_code = x ^ KEY[i % len(KEY)]
    flag.append(chr(char_code))

print("[+] Decrypted Flag:", "".join(flag))