import hashlib
import hmac
import math
import struct
def mystery(x):
# Accept either str or number; normalize deterministically to bytes
if isinstance(x, (int, float)):
# Canonicalize floats/ints via IEEE-754 double bytes so many "similar" values differ.
# Reject NaN/inf to avoid weirdness.
if isinstance(x, float) and (math.isnan(x) or math.isinf(x)):
return False
b = struct.pack(">d", float(x))
elif isinstance(x, str):
b = x.encode("utf-8")
else:
return False
# A tiny, fixed "key" derived from a non-obvious arithmetic identity (still constant).
key = (str(sum(i*i*i for i in range(1, 11))) + ":" + str(math.prod(range(1, 8)))).encode("ascii")
# Build a MAC, then hash again; check for a fairly rare pattern in the digest.
mac = hmac.new(key, b, hashlib.sha256).digest()
d = hashlib.sha256(mac + b"|" + hashlib.sha1(b).digest()).hexdigest()
# Condition: digest must begin with 5 zero hex chars and end with exactly '42'
return d.startswith("00000") and d.endswith("42")
def mystery(x):
# Input must be a string of length 7
if not isinstance(x, str) or len(x) != 7:
return False
# Convert to ASCII values
v = [ord(c) for c in x]
# 1. Boundary constraints
# The string must start with 'P' (80) and end with '!' (33)
if v[0] != 0x50 or v[6] != 0x21:
return False
# 2. Linear constraints
# The sum of all ASCII values must be 683
if sum(v) != 683:
return False
# The alternating sum (c0 - c1 + c2 - c3 + c4 - c5 + c6) must be 3
# This effectively creates a relation between the even and odd indexed characters.
alt_sum = sum(v[i] * (1 if i % 2 == 0 else -1) for i in range(7))
if alt_sum != 3:
return False
# 3. Structural constraints
# The 3rd and 4th characters must be identical
if v[2] != v[3]:
return False
# The 2nd and 5th characters must satisfy a specific bitwise XOR relationship
if v[1] ^ v[4] != 0x19: # 0x19 is 25
return False
# 4. Filter constraints to ensure uniqueness
# The 2nd character must be a lowercase vowel
if x[1] not in "aeiou":
return False
# The 3rd character must be lexically greater than 'x' (ASCII 120)
if v[2] <= 120:
return False
return True
import math
def mystery(x):
# Only accept strings
if not isinstance(x, str):
return False
# Exact length requirement
if len(x) != 22:
return False
# Must be printable ASCII (space..~)
if any(not (32 <= ord(c) <= 126) for c in x):
return False
# Character-class constraints to narrow the search space without making it obvious
if not (x[:4].isalpha() and x[:4].isupper()):
return False
if x[4] != "-":
return False
if not x[5:9].isdigit():
return False
if x[9] != "-":
return False
if not x[10:14].islower():
return False
if x[14] != "-":
return False
if not x[15:22].isalnum():
return False
# Nonlinear arithmetic constraints (hard to solve by inspection)
v = [ord(c) for c in x]
# A few "checksum-like" relations mixing positions
if (sum(v[i] * (i + 1) for i in range(22)) % 997) != 827:
return False
if (sum((v[i] ^ v[21 - i]) for i in range(11)) % 256) != 34:
return False
# A disguised exact match using floating-point edge conditions but stable for integers
t = sum(((v[i] + 17) * (v[i] - 31)) for i in range(22))
if not math.isfinite(float(t)):
return False
if (t % 10007) != 540:
return False
# Final gate: a single exact string, but masked behind the above constraints
target = "NOVA-0420-lumen-A1b2C3d"
return x == target
def mystery(x):
# Input validation: must be an 8-character string
if not isinstance(x, str) or len(x) != 8:
return False
# Convert string to list of ASCII integers
v = [ord(c) for c in x]
# Constraint 1: Structural anchors
# The 5th character must be '_' (underscore)
# The 8th character must be '!' (exclamation mark)
if v[4] != 0x5F or v[7] != 0x21:
return False
# Constraint 2: Aggregate checksums
# Simple sum must be 619
if sum(v) != 619:
return False
# Weighted sum must be 2460
# Formula: v[0]*1 + v[1]*2 + ... + v[7]*8
if sum(v[i] * (i + 1) for i in range(8)) != 2460:
return False
# Constraint 3: Bitwise correlation
# Treat the first 4 characters as one 32-bit integer (big endian)
# Treat the last 4 characters as another 32-bit integer
part_a = (v[0] << 24) | (v[1] << 16) | (v[2] << 8) | v[3]
part_b = (v[4] << 24) | (v[5] << 16) | (v[6] << 8) | v[7]
# Their XOR sum must equal this specific magic number
# 475470916 is 0x1C575C44 in hex
if (part_a ^ part_b) != 475470916:
return False
# Constraint 4: Nonlinear relation
# A specific algebraic relationship between the first four characters
if (v[0] * v[2] + v[1] * v[3]) != 17911:
return False
return True
import hashlib
import math
import struct
def mystery(x):
# Accept str or number; canonicalize to bytes
if isinstance(x, str):
b = x.encode("utf-8")
tag = b"S"
elif isinstance(x, (int, float)):
if isinstance(x, float) and (math.isnan(x) or math.isinf(x)):
return False
# Canonical IEEE-754 double representation
b = struct.pack(">d", float(x))
tag = b"N"
else:
return False
# Cheap decoys to waste time (but deterministic)
h1 = hashlib.sha256(b"decoy|" + tag + b + b"|1").digest()
h2 = hashlib.blake2b(b"decoy|" + h1 + b"|2", digest_size=32).digest()
_ = int.from_bytes(h2[:8], "big") ^ int.from_bytes(h2[8:16], "big")
# The real gate: a "public-key looking" check that is actually just one exact preimage.
target = b"paper-boat/13"
d = hashlib.sha256(b"gate|" + target).digest()
return hashlib.sha256(b"gate|" + b).digest() == d
import functools
def mystery(x):
# Validate container format: Must look like "Key{...}"
if not isinstance(x, str) or not x.startswith("Key{") or not x.endswith("}"):
return False
content = x[4:-1]
# Enforce lowercase hex characters and commas only (no 0x prefix, no uppercase)
if not set(content).issubset(set("0123456789abcdef,")):
return False
parts = content.split(",")
# Must contain exactly 5 numbers
if len(parts) != 5:
return False
try:
# Parse hex strings to integers
v = [int(p, 16) for p in parts]
except ValueError:
return False
# Constraint 1: The sequence must be strictly increasing
if any(v[i] >= v[i+1] for i in range(4)):
return False
# Constraint 2: The first three numbers form an Arithmetic Progression (AP)
# a, a+d, a+2d => (a) + (a+2d) == 2*(a+d)
if v[0] + v[2] != 2 * v[1]:
return False
# Constraint 3: The last three numbers form a Geometric Progression (GP)
# b, b*r, b*r^2 => b * b*r^2 == (b*r)^2
if v[2] * v[4] != v[3] ** 2:
return False
# Constraint 4: The difference between the last and first number is fixed
if v[4] - v[0] != 121:
return False
# Constraint 5: The product of all five numbers must match this large constant
product = functools.reduce(lambda a, b: a * b, v)
if product != 69574032:
return False
return True
import hashlib
import struct
import math
def mystery(x):
# normalize to bytes, but with a twist for numbers
if isinstance(x, str):
b = x.encode("utf-8")
kind = 0x53 # 'S'
elif isinstance(x, (int, float)):
if isinstance(x, float) and (math.isnan(x) or math.isinf(x)):
return False
# canonical IEEE-754 bytes for floats; ints go through float() too
b = struct.pack(">d", float(x))
kind = 0x4E # 'N'
else:
return False
# "looks like crypto": chained hashes with mixed domains
h = hashlib.sha256(b"\x01salt|" + bytes([kind]) + b).digest()
h = hashlib.blake2s(b"\x02mix|" + h + b[::-1], digest_size=32).digest()
h = hashlib.sha3_256(b"\x03fin|" + h + b + bytes([len(b) & 0xFF])).digest()
# derive a seemingly-arbitrary 64-bit check value
chk = int.from_bytes(h[:8], "big") ^ int.from_bytes(h[8:16], "big")
# final gate: forced equality to an embedded value, but disguised behind chk
# (so preimage hunting via chk is still effectively "find the exact b")
target = b"steganography>entropy"
want = int.from_bytes(hashlib.sha3_256(b"\x03fin|" +
hashlib.blake2s(
b"\x02mix|" +
hashlib.sha256(b"\x01salt|" + b"\x53" + target).digest() +
target[::-1],
digest_size=32
).digest() +
target + bytes([len(target) & 0xFF])
).digest()[:8], "big") ^ \
int.from_bytes(hashlib.sha3_256(b"\x03fin|" +
hashlib.blake2s(
b"\x02mix|" +
hashlib.sha256(b"\x01salt|" + b"\x53" + target).digest() +
target[::-1],
digest_size=32
).digest() +
target + bytes([len(target) & 0xFF])
).digest()[8:16], "big")
return chk == want
def mystery(x):
# Input must be an 8-character string
if not isinstance(x, str) or len(x) != 8:
return False
v = [ord(c) for c in x]
# Constraint 1: Global checksum
# The sum of all ASCII values must be 688
if sum(v) != 688:
return False
# Constraint 2: Pairwise Sum and XOR relations
# We treat the string as 4 pairs of characters.
# For each pair (a, b), we check their sum and their XOR.
# Note: For any integers a and b, knowing (a+b) and (a^b) strictly constrains the values,
# though swapping a and b is possible, and sometimes other bit-shifts are possible.
targets = [
(178, 44), # Pair 0: v[0], v[1]
(201, 1), # Pair 1: v[2], v[3]
(201, 41), # Pair 2: v[4], v[5]
(108, 6) # Pair 3: v[6], v[7]
]
for i in range(4):
a, b = v[2*i], v[2*i+1]
t_sum, t_xor = targets[i]
if a + b != t_sum or a ^ b != t_xor:
return False
# Constraint 3: Specific filters to resolve ambiguity
# These ensure there is only one valid string among the candidates generated above.
# Filter for Pair 0
if v[0] % 5 != 2:
return False
# Filter for Pair 1
if v[2] >= v[3]:
return False
# Filter for Pair 2
if v[5] % 5 != 1:
return False
# Filter for Pair 3
if abs(v[6] - v[7]) != 6 or v[6] >= v[7]:
return False
return True
import hashlib
import struct
import math
def mystery(x):
# normalize input -> bytes (numbers use IEEE-754 double, strings use UTF-8)
if isinstance(x, str):
b = x.encode("utf-8")
kind = b"S"
elif isinstance(x, (int, float)):
if isinstance(x, float) and (math.isnan(x) or math.isinf(x)):
return False
b = struct.pack(">d", float(x))
kind = b"N"
else:
return False
# compute a "fingerprint" that looks like it depends on many properties
h = hashlib.blake2s(b"\x01" + kind + b, digest_size=32).digest()
h = hashlib.sha3_256(b"\x02" + h + bytes([len(b) & 0xFF]) + b[::-1]).digest()
tag = hashlib.sha256(b"\x03" + h + b"\x00" + kind).hexdigest()
# seemingly arbitrary constraint (rare prefix/suffix), plus an embedded equality trap
if not (tag.startswith("0e") and tag.endswith("c0")):
return False
# final gate: forces the exact intended preimage, but hidden behind the earlier checks
target = b"brittle/maze::v2|17"
want = hashlib.sha256(b"\x03" + hashlib.sha3_256(
b"\x02" + hashlib.blake2s(b"\x01" + b"S" + target, digest_size=32).digest()
+ bytes([len(target) & 0xFF]) + target[::-1]
).digest() + b"\x00" + b"S").hexdigest()
return tag == want
def mystery(x):
# Input must be an 8-character string
if not isinstance(x, str) or len(x) != 8:
return False
v = [ord(c) for c in x]
# Interpret the string as two 32-bit big-endian integers
# 'a' comes from the first 4 characters
# 'b' comes from the last 4 characters
a = (v[0] << 24) | (v[1] << 16) | (v[2] << 8) | v[3]
b = (v[4] << 24) | (v[5] << 16) | (v[6] << 8) | v[7]
# Constraint 1: The two integers must satisfy this XOR relation
if (a ^ b) != 0x6E5B5644:
return False
# Constraint 2: The two integers must satisfy this arithmetic Sum relation
if (a + b) != 0x70A39686:
return False
# Constraint 3: Fixed anchor character
if x[4] != '-':
return False
# Constraint 4: Polynomial Rolling Hash to verify the permutation
# Base 31, Modulo 10^9 + 7
h = 0
MOD = 1000000007
for c in v:
h = (h * 31 + c) % MOD
if h != 815623291:
return False
return True