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NoorQureshiNoorQureshi/SploitAgent/skills/ctf/ctf-crypto/SKILL.md

ctf-crypto

Skill26 starsChanged 32 days agoMITView on GitHub

CTF crypto challenge playbook — read the provided encryption script like a spec, attack RSA parameter weaknesses, XOR/multi-time-pad oracles, and homemade ciphers with math tooling (z3, sympy, sage). Load when the handout is .py/.sage math, n/e/c values, or an "encryption service" on nc. Signals: "crypto" category, chall.py with Crypto.Util.number, e=3 or e=65537, "encrypted_flag", XOR with a key, nc service that encrypts your input.

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The file

We split this skill into the rules it gives an agent

2 rules. 0 are common, 0 are distinctive, and 2 are very specific to this project.

  • 0

    Common

    Most of the 3 edited versions we read keep these.

  • 0

    Distinctive

    This author's own choices. Worth a look.

  • 2

    Very specific

    Tied to their project's names, files or tools. Probably skip.

  • Common
  • Distinctive
  • Very specific
  1. Very specific:

    Ask: *is every step invertible given the key?

    line 72
  2. Very specific:

    The printed "leak" is intentional: If the script hands you hint = p ^ q or leak = d & ((1<<512)-1), that's the solve path — stop looking …

    line 80

“Common” is measured: kept by at least 6 in 10 of the 3 edited versions of this file we could read. “Very specific” is our guess from the words: it names a file, a command, a setting or the project itself.

Keep what fits and drop the rest. Your version, ready for your agents.
Files
SKILL.md

CTF crypto challenges

When it applies

The handout is a short encryption script (chall.py, rarely sage), a set of numbers (n, e, c / intercepted ciphertexts), or an nc service that encrypts/signs things for you. This skill covers the CTF shapes — reading the script for the flaw, XOR games, homemade ciphers, and math tooling. For the deep attack mechanics it defers to crypto-rsa-attacks and crypto-oracle-attacks; run the RSA battery there first.

Why it works

CTF crypto is a code review against math: the script is short, self-contained, and contains exactly one mistake the author planted — a reused nonce, a factorable n, an invertible homegrown round function, or an oracle the service hands you. Find the one line that looks "clever" and the challenge is usually done; the rest is algebra.

Method

  1. Read the script as a spec, top to bottom. Note exactly what's given (printed values) and what's hidden (keys, seeds, flag). The flaw is almost always in: key/nonce generation (random instead of secrets, timestamp seeds, getPrime with small bits, reused k), the composition (same key twice, encrypt-then-leak-something), or the padding (raw RSA, ASCII armor that leaks length).
  2. Triage encodings before crypto. Half of "crypto" challenges are encoding stacks: base64/32/ 85, hex, morse, brainfuck, base-N with a custom alphabet, repeated base64. CyberChef's magic wand or a quick decode loop settles these in minutes — don't reach for sage on a base85.
  3. RSA → run the known battery. Extract n, e, c; hand the key to RsaCtfTool + factordb, then check the CTF-specific shapes explicitly (crypto-rsa-attacks for mechanics):
    • Multiple keys/ciphertexts in one file → gcd pairwise (shared prime), common modulus (same n, two e), or Håstad broadcast (same message, small e, several n).
    • p and q generated from a weak PRNG / shared high bits / q = next_prime(p) → Fermat or regenerate the PRNG stream.
    • Extra printed values (d mod (p-1), p+q, dp, partial bits of p) → Coppersmith-style recovery (sage) or direct algebra; these "hint leaks" are the intended path.
  4. XOR challenges.
    • Single-byte XOR: brute all 256 keys, score by English frequency (xortool automates).
    • Repeating-key XOR: guess key length by Hamming distance / index of coincidence, solve each column as single-byte.
    • Multi-time pad (same keystream XORs several messages, classic otp reuse): crib-drag — XOR two ciphertexts and guess words ("the ", "flag{") at each position; each confirmed crib extends the keystream.
    • XOR with a known plaintext anywhere (a known header, the flag format flag{) recovers keystream bytes directly.
  5. Encryption services are oracles — script them. nc service that encrypts your input:
    • Your input prepended to the flag, ECB → byte-at-a-time flag recovery (crypto-oracle-attacks).
    • Same plaintext always → same ciphertext: codebook lookup / chosen-plaintext mapping.
    • Decrypt-anything-but-the-flag → malleability games: bit-flip CBC, RSA blinding (c·r^e mod n, unblind the result).
    • It always helps to send structured probes: all-zero blocks, single-byte deltas, length sweeps (leaks block size and mode).
  6. Homegrown ciphers → model, don't guess. Custom rounds (add-rotate-xor, matrix mods, LFSRs, "Feistel-ish"): symbolically execute or invert them.
    • Linear operations over GF(2) → write as linear equations, solve with z3 or sage.
    • Small state (LFSR) → Berlekamp–Massey from a known keystream.
    • S-box + permutation networks → differential or just brute the tiny key space the author used.
    • Ask: is every step invertible given the key? Then the only question is the key space — brute 16/24/32-bit keys before doing anything clever.
  7. Bring the math tools. z3 for constraint-shaped problems (flag bytes satisfying check equations); sympy for modular algebra and small discrete logs; sage for lattices (LLL / Coppersmith / hidden-number problems); long_to_bytes/bytes_to_long endianness discipline throughout.

Gotchas

  • The printed "leak" is intentional. If the script hands you hint = p ^ q or leak = d & ((1<<512)-1), that's the solve path — stop looking for a second bug.
  • Python random is not crypto — random.seed(time.time()) or Mersenne Twister output visible anywhere → clone the state (624 consecutive 32-bit outputs fully recover MT19937).
  • e and phi mistakes — wrong totient (using n for (p-1)(q-1)) makes a key that looks right but fails; verify by decrypting a known value (crypto-rsa-attacks).
  • Multi-prime RSA (n = pqr) — factorization tools still work; remember phi = (p-1)(q-1)(r-1).
  • Encoding is not encryption — if "ciphertext" decodes cleanly to printable text, keep peeling encodings before assuming a cipher.
  • Byte-vs-int conversion bugs in your solver — leading zero bytes vanish in bytes_to_long; pad the recovered plaintext back to the expected length.

Verify success

The recovered plaintext contains the flag in the event's format, and you can re-derive it deterministically (script reproduces the key/plaintext from the given artifacts, not from luck).

References

Mechanics: crypto-rsa-attacks (factordb, Fermat, Wiener, Håstad), crypto-oracle-attacks (padding oracle, ECB byte-at-a-time, length extension); tooling: z3, sage (LLL/Coppersmith), xortool, CyberChef. Triage via ctf-methodology.

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