noname_User's Unbreakable Python? My Custom Obfuscation Engine

Author:
noname_User

Language:
Python

Upload:
2026-02-23 07:49

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Platform:
Windows

Difficulty:
2.0

Quality:
4.0

Arch:
other

Downloads:
386

Size:
5.95 KB

Writeups:
0

Comments:
2

Description

I created this custom obfuscation engine for Python to test reverse engineering skills. It features a unique 7-layer protection: Obfuscated Shell (Indirect calls) Hybrid RSA + AES-256 (Key derivation) Chaos VM (Hardware-bound logic) Anti-Analysis (Anti-Debug & Anti-VM) Integrity Self-Check (SHA-256) MBA & Junk Code Control Flow Flattening (CFF) The Goal: Can you bypass all layers and find the core logic? Note: This is my original work. No commercial protectors were used. Language : python Input: No input required Output: "An RGB 'noname' text effect in the console.

Labels ?

Anti-debugging String / data encryption XOR Base64 (encryption) AES (encryption) Packer MPRESS Crypto / hash algorithm Base64 (crypto) RSA AES (crypto) SHA-256 Code virtualization / VM Control-flow obfuscation Control-flow flattening (CFF) Anti-disassembly Junk / garbage bytes Anti-tamper / integrity Custom / generic obfuscation Anti-VM / sandbox

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CrackNotMe on 2026-02-24 19:02: Layer 1 and 2 Extracted a Base64 and Zlib encoded payload that launches a hidden daemon thread named saadjqymaqfc to run all protections in the background. Layer 3 and 4 Chaos VM and Hardware Fingerprint Found Anti-VM checks scanning for MAC addresses like 080027, 000569, 000c29, 001c14, 005056. Extracted the author's target machine SHA256 hardware fingerprint: 90178d2d1e81e3cc1373ae36327277909ea0a904108c26fb80fec88755553bbd which binds to WMI disk, BIOS, and CPU serials. Extracted the Chaos algorithm math logic which uses 50 iterations of the logistic map: 3.99 * val * (1.0 - val). Recovered the final derived Master Key: 64322080736143896125652295414509504119508758357723799657495947910577489034260 Layer 3 Cryptography Found the core payload hidden in a massive Base85 string. Bypassed the decrypter which uses large integer modular exponentiation (pow function) and a custom XOR routine disguised as addition and subtraction math. Layer 6 Integrity Self-Check Extracted the expected SHA256 checksum of the unmodified file: 9de1a7e0b2d79d9be0d1f5b71bee50ce2656856c4b78f05c60d4ad242b0eaf0c Layer 7 Bytecode Obfuscation and Payload Extracted the final hidden function named: smooth_rgb_normal_text Discovered the obfuscator spoofed the bytecode argument count to 0 to crash debuggers, when it actually requires 4 arguments. Extracted the final hardcoded variables for the effect: text is noname, spacing 1, speed 0.1, wave_width 0.5, and sine wave RGB multipliers using 127, 128, 2, and 4. Identified the intentional silent anti-analysis tactic relying on compiled Python 3.10 bytecode opcodes that intentionally segfault newer Python versions like 3.13 if not cleanly extracted.

plikan on 2026-04-10 14:39: source code: import sys import math import time import os def smooth_rgb_normal_text(): text = "noname" spacing = 0.5 speed = 0.1 wave_width = 2.0 sys.stdout.write("\033[?25l") try: for t in range(0, 1000): output = [] for i, char in enumerate(text): freq = wave_width * i r = int((math.sin(freq + t * speed) + 1) / 2 * 255) g = int((math.sin(freq + 2.0 + t * speed) + 1) / 2 * 255) b = int((math.sin(freq + 4.0 + t * speed) + 1) / 2 * 255) output.append(f"\033[38;2;{r};{g};{b}m{char}\033[0m") output.append(" " * int(spacing)) sys.stdout.write("\r" + "".join(output)) sys.stdout.flush() time.sleep(0.05) except KeyboardInterrupt: sys.stdout.write("\033[?25h") sys.stdout.write("\n") if __name__ == "__main__": if os.name == "nt": os.system("cls") else: os.system("clear") smooth_rgb_normal_text()

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noname_User's Unbreakable Python? My Custom Obfuscation Engine