reversefreehard

loicense

pingCTF

Task: License keygen for SmartAttend ELF with DWARF debug info, validates keys via 10x10 linear system. Solution: Reverse key format (11-char metadata + 10 IEEE-754 floats as hex), solve linear system from HWID ASCII values, bruteforce metadata for numerical stability.

$ ls tags/ techniques/
linear_system_solvingieee754_encodingmetadata_bruteforce

loicense - pingCTF

Description

License validation challenge with SmartAttend ELF binary.

Given an ELF binary SmartAttend with DWARF debug info and a remote checker that sends 100 random HWIDs. For each HWID, generate a valid license key. Pass all 100 rounds to get the flag.

Analysis

Binary Structure

The binary is a 64-bit ELF with full DWARF debug symbols, making reverse engineering straightforward with radare2:

SmartAttend: ELF 64-bit LSB executable, x86-64, with debug_info, not stripped

Key Format (from decode_signature_from_key)

License key structure (91 characters total):

  • 11 chars: Metadata prefix (alphanumeric: 0-9A-Z)
  • 80 chars: 10 IEEE-754 float32 values encoded as 8 hex digits each

HWID Structure

The 99-character HWID (charset 0-9A-Z) is split into 9 rows of 11 characters each.

Validation Model (from compute_system_loss / probe_signature_candidate)

The validation constructs a 10x10 linear system:

  1. Each HWID row provides coefficients: first 10 chars = coefficients, 11th char = RHS
  2. The metadata row (from key) provides the 10th equation
  3. Coefficients are raw ASCII values of characters (e.g., 'A' = 65, '0' = 48)
  4. The 10 float values from the key must satisfy all 10 equations

The system validates: A @ signature = b where:

  • A is 10x10 matrix from ASCII values of first 10 chars per row
  • b is 10-element vector from ASCII values of 11th char per row
  • signature is the 10 float values decoded from the key

Solution

Keygen Strategy

  1. Parse HWID into 9 rows of 11 ASCII values each
  2. Try multiple metadata candidates to find one that yields a well-conditioned 10x10 system
  3. Solve the linear system using numpy
  4. Encode solution floats as IEEE-754 hex
  5. Select metadata with smallest float32 residual for numerical stability

Solve Script

#!/usr/bin/env python3 import struct import numpy as np def build_key(hwid: str, meta_candidates: list[str]) -> tuple[str, float]: """Generate license key for given HWID.""" # Parse HWID into 9 rows of ASCII values rows = [ np.array([float(ord(c)) for c in hwid[i*11:(i+1)*11]], dtype=np.float64) for i in range(9) ] base_matrix = np.vstack([row[:10] for row in rows]) base_rhs = np.array([row[10] for row in rows], dtype=np.float64) best_key, best_residual = None, float("inf") for meta in meta_candidates: # Add metadata as 10th row meta_row = np.array([float(ord(c)) for c in meta], dtype=np.float64) matrix = np.vstack([base_matrix, meta_row[:10]]) rhs = np.append(base_rhs, meta_row[10]) try: signature = np.linalg.solve(matrix, rhs).astype(np.float32) except np.linalg.LinAlgError: continue # Check residual after float32 rounding residual = float(np.max(np.abs(matrix @ signature.astype(np.float64) - rhs))) if residual < best_residual: # Encode floats as IEEE-754 hex key = meta + "".join( f"{struct.unpack('>I', struct.pack('>f', float(v)))[0]:08X}" for v in signature ) best_key, best_residual = key, residual if residual < 1e-6: break return best_key, best_residual

Remote Interaction

Connect to remote, receive 100 HWIDs, generate and send keys:

for round_idx in range(100): hwid = recv_hwid(sock) # 99-char alphanumeric key, residual = build_key(hwid, META_CANDIDATES) sock.sendall((key + "\n").encode())

Metadata Bruteforce

To ensure numerical stability across 100 random HWIDs, generate ~500 metadata candidates and pick the one with smallest residual:

def make_meta_candidates() -> list[str]: rng = random.Random(1337) alphabet = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ" seeds = ["10000000000", "1234567890A", ...] for _ in range(500): seeds.append(rng.choice("0123456789") + "".join(rng.choice(alphabet) for _ in range(10))) return seeds

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