ctf-crypto

Tool passed security scan

Risky tool definition detected

```python for attempt in range(3000): r.sendlineafter(b"prompt: ", b"00" * 8) # Fixed zero response result = r.recvline() if b"successful" in result: break ``` --- ## Deterministic OTP with Load-Balanced Backends (Pragyan 2026) **Pattern (DumCows):** Service encrypts data with deterministic keystream that resets per connection. Multiple backends with different keystreams behind a load balancer.

Malicious tool definition detected

Risky tool definition detected

Tool passed security scan

Risky tool definition detected

Risky tool definition detected

Tool: ctf-crypto/modern-ciphers.md [2/2] Description: aug[col], aug[pivot] = aug[pivot], aug[col] for r in range(64): if r != col and aug[r][col]: aug[r] = [a ^ b for a, b in zip(aug[r], aug[col])] return [row[64:] for row in aug] ``` **Centroid clustering for key classification:** ```python # For each bit position, measure bias at two output positions # 4 possible (k0[i], k1[i]) pairs → 4 centroid patterns # Uses sign-pattern mask CMASK=0x73 to account for bit-position-dependent behavior # Clas

Tool passed security scan

Risky tool definition detected

n = p * q # = p^2 e = 65537 wrong_phi = (p - 1) * (q - 1) # = (p-1)^2 d = inverse(e, wrong_phi) # passes server validation # Server encrypts flag with our key, test decryption fails → leaks ciphertext c # Decrypt with correct totient: real_phi = p * (p - 1) real_d = inverse(e, real_phi) flag = pow(c, real_d, n) ``` **Key insight:** `phi(p^2) = p*(p-1)`, NOT `(p-1)^2`. When a server validates RSA parameters but uses `(p-1)*(q-1)` without checking `p != q`, setting `p=q` creates a working key that

Tool passed security scan

Context Poisoning

Risky tool definition detected

Tool passed security scan

Context Poisoning

Guardrail Bypass