def mystery(x):
# Accept either a string or a number.
if isinstance(x, str):
b = x.encode('utf-8')
val = int.from_bytes(b, 'big')
elif isinstance(x, (int, float)):
try:
val = int(x)
except Exception:
return False
else:
return False
# A few obfuscated numerical checks (all must hold).
# These combine linear congruences with different moduli and a small residue check.
a = (val * 17 + 1000) % 1000003
b = (val * 31337 + 271828) % 1000000007
c = val % 97
return a == 311879 and b == 573331547 and c == 51
import zlib
import numbers
def mystery(x):
"""
This function subjects the input to a series of transformations and checks.
Finding an input that passes all checks is the challenge.
"""
# The input must be a string of a specific length. This is the first gate.
if not isinstance(x, str) or len(x) != 13:
return False
try:
# The core of the puzzle involves a specific data transformation pipeline.
# 1. The string is encoded to bytes.
b = x.encode('utf-8')
# 2. The bytes are compressed. The choice of compression adds a
# non-linear transformation that is difficult to reverse-engineer.
compressed = zlib.compress(b)
# 3. The resulting compressed bytes are interpreted as a single large integer.
val = int.from_bytes(compressed, 'big')
# To pass, this integer must satisfy two non-obvious modular arithmetic conditions.
# These checks act like a checksum, but are much harder to reverse than a simple hash.
# The moduli are large prime numbers.
cond1 = (val % 1000000007) == 495345262
# A bit shift is added before the second check to further obfuscate the relationship
# between the two conditions and the original integer `val`.
cond2 = ((val >> 32) % 998244353) == 415514787
return cond1 and cond2
except Exception:
# Gracefully handle any errors during the transformation process.
return False
import hashlib
import zlib
import binascii
def mystery(x):
"""
Accepts a string or number. Performs several non-obvious checks
(hashes, checksums, compression artifact) on the UTF-8 bytes of the input.
Very few inputs will satisfy all of them.
"""
# Normalize numbers to strings; everything else must be a string.
if isinstance(x, (int, float)):
x = str(x)
if not isinstance(x, str):
return False
b = x.encode('utf-8')
# A set of independent, opaque checks using well-known algorithms.
# (These are deliberately chosen so that only a very specific byte sequence
# will pass every test.)
if hashlib.sha256(b).hexdigest() != (
"e3b0c44298fc1c149afbf4c8996fb924"
"27ae41e4649b934ca495991b7852b855"
):
return False
if hashlib.md5(b).hexdigest() != "d41d8cd98f00b204e9800998ecf8427e":
return False
# Standard CRC/Adler checks
if binascii.crc32(b) != 0:
return False
if zlib.adler32(b) != 1:
return False
# The exact compressed-bytes artifact for the empty input under zlib.compress
if zlib.compress(b) != b'x\x9c\x03\x00\x00\x00\x00\x01':
return False
# If all opaque checks pass, accept.
return True
import base64
import struct
def mystery(x):
"""
This puzzle requires a very specific structured input.
The input must be a string that decodes and unpacks into a sequence
of numbers satisfying a hidden mathematical property.
"""
# The input must be a string.
if not isinstance(x, str):
return False
try:
# It must be a valid base64-encoded string.
# The `validate=True` option rejects non-alphabet characters,
# providing an early and subtle hint about the input's nature.
b = base64.b64decode(x, validate=True)
except (ValueError, TypeError):
return False
# The decoded bytestring must have a very specific length. This is a
# crucial structural clue.
if len(b) != 24:
return False
# Unpack the bytestring into a tuple of six 32-bit unsigned integers.
# The big-endian format ('>') is an arbitrary but fixed choice that
# must be discovered or guessed by the solver.
nums = struct.unpack('>6I', b)
# First check: A hidden recurrence relation must hold for the numbers.
# This is a Linear Congruential Generator (LCG). The constants are
# taken from a common C library implementation, which is not obvious.
# A brute-force attack on the numbers is hard because the search space is huge.
for i in range(5):
expected_next = (nums[i] * 1664525 + 1013904223) & 0xFFFFFFFF
if nums[i+1] != expected_next:
return False
# Second check: To prevent finding *any* valid LCG sequence, the sum
# of the numbers must match a specific, seemingly arbitrary "magic" value.
# This effectively pins down the sequence to one generated from a specific seed.
if sum(nums) != 11286863709:
return False
# If all checks pass, we've found the needle in the haystack.
return True
def mystery(x):
"""
Accepts either an integer/float or a numeric string. The input is converted to an int
and must satisfy many independent, seemingly unrelated constraints (digit-pattern,
digit-sum, many modular congruences, and a low-16-bit check). All must hold.
"""
# Normalize numeric inputs to int, accept numeric strings as well.
try:
if isinstance(x, (int, float)):
val = int(x)
elif isinstance(x, str):
# allow an optional leading plus sign for strings like "+123"
if x.startswith('+') and x[1:].isdigit():
val = int(x)
elif x.isdigit():
val = int(x)
else:
return False
else:
return False
except Exception:
return False
s = str(val)
# Structural/digit checks (easy to state, but combined with the arithmetic checks below
# they make the set of valid inputs tiny).
if len(s) != 10:
return False
if not s.startswith('1234'):
return False
if not s.endswith('7890'):
return False
if sum(int(d) for d in s) != 45:
return False
# A collection of modular congruences (all must hold).
mods = {
3: 0,
4: 2,
5: 0,
7: 3,
8: 2,
9: 0,
11: 5,
13: 10,
17: 10,
19: 7,
23: 18,
29: 16,
31: 20,
}
for m, r in mods.items():
if val % m != r:
return False
# A low-16-bit pattern check (the bottom 16 bits must equal this specific value).
if (val & 0xFFFF) != 722:
return False
return True
def mystery(x):
"""
This puzzle requires finding a specific graph structure, encoded as a string.
The graph's properties are tested using matrix exponentiation, representing
the number of walks of a certain length.
"""
if not isinstance(x, str):
return False
# The graph must have exactly 7 vertices, labeled 'A' through 'G'.
labels = "ABCDEFG"
label_to_idx = {label: i for i, label in enumerate(labels)}
n = len(labels)
adj = [[0] * n for _ in range(n)]
try:
# Parse the edge list from the input string.
# The format must be "U-V,W-X,..." with specific constraints.
edges = set()
if not x or '-' not in x:
return False
parts = x.split(',')
for part in parts:
nodes = part.split('-')
if len(nodes) != 2: return False
u_label, v_label = nodes[0].strip(), nodes[1].strip()
# Vertex labels must be valid and distinct. No self-loops.
if u_label not in label_to_idx or v_label not in label_to_idx: return False
if u_label == v_label: return False
u, v = label_to_idx[u_label], label_to_idx[v_label]
# Canonical edge representation (u < v) to handle "A-B" vs "B-A"
# and prevent duplicate edges.
edge = tuple(sorted((u, v)))
if edge in edges: return False
edges.add(edge)
# Populate the adjacency matrix for an undirected graph.
adj[u][v] = 1
adj[v][u] = 1
except Exception:
return False
# The core mathematical check.
# We are looking for a graph where the total number of walks of length 5
# is a specific "magic" number. This number is the sum of all entries
# in the 5th power of the adjacency matrix.
def _matmul(A, B):
C = [[0] * n for _ in range(n)]
for i in range(n):
for j in range(n):
for k in range(n):
C[i][j] += A[i][k] * B[k][j]
return C
def _matpow(A, k):
res = [[int(i==j) for j in range(n)] for i in range(n)] # Identity matrix
base = A
while k > 0:
if k % 2 == 1: res = _matmul(res, base)
base = _matmul(base, base)
k //= 2
return res
try:
adj_pow_5 = _matpow(adj, 5)
# The sum of all elements in A^5 must equal the magic number.
total_walks = sum(sum(row) for row in adj_pow_5)
return total_walks == 394
except Exception:
return False
def mystery(x):
"""
Accept either an integer/float or a numeric string. The input is normalized to an int
and must pass a dense collection of independent checks (digit structure, digit-sum,
multiple modular residues, and low-bit patterns). All checks must hold.
"""
# Normalize and validate type
try:
if isinstance(x, (int, float)):
val = int(x)
elif isinstance(x, str):
# allow optional leading plus sign
if x.startswith('+') and x[1:].isdigit():
val = int(x)
elif x.isdigit():
val = int(x)
else:
return False
else:
return False
except Exception:
return False
s = str(val)
# Structural digit constraints (easy to state but restrictive in combination)
if len(s) != 8:
return False
if not s.startswith('2026'):
return False
if not s.endswith('17'):
return False
if sum(int(d) for d in s) != 19:
return False
if len(set(s)) != 5:
return False
# Collection of modular congruences (all must hold)
if val % 2 != 1: # odd
return False
if val % 3 != 1:
return False
if val % 4 != 1:
return False
if val % 5 != 2:
return False
if val % 7 != 3:
return False
if val % 8 != 5:
return False
if val % 9 != 1:
return False
if val % 11 != 2:
return False
if val % 101 != 22:
return False
# Low-bit and trailing-digit checks
if (val & 0xFFFF) != 9493:
return False
if (val & 0xFF) != 117:
return False
if int(s[-3:]) != 117:
return False
# A slightly redundant derived check to reduce accidental collisions.
# Sort the digits and form an integer from them (leading zeros allowed in sort).
sorted_digits = ''.join(sorted(s))
if int(sorted_digits) != 112267: # int('00112267') == 112267
return False
return True
import hashlib
def mystery(x):
"""
This puzzle challenges you to find the initial state of a cellular automaton
that evolves into a specific, hidden final state after a set number of steps.
"""
if not isinstance(x, str):
return False
# The grid is 10x10 and wraps around (a toroidal array).
GRID_SIZE = 10
# The automaton has three states: 0 (Off), 1 (Firing), 2 (On).
grid = [[0] * GRID_SIZE for _ in range(GRID_SIZE)]
try:
# The input string `x` defines the initial "On" cells.
# The required format is "row1,col1;row2,col2;...".
initial_on_cells = set()
if x: # An empty string corresponds to an empty initial state.
parts = x.split(';')
for part in parts:
coords = part.split(',')
if len(coords) != 2: return False
r, c = int(coords[0]), int(coords[1])
if not (0 <= r < GRID_SIZE and 0 <= c < GRID_SIZE): return False
initial_on_cells.add((r, c))
# To pass, the initial pattern must contain exactly 7 "On" cells.
# This is a crucial constraint that defines the search space.
if len(initial_on_cells) != 7:
return False
for r, c in initial_on_cells:
grid[r][c] = 2 # State 2 is "On"
except (ValueError, TypeError):
return False
# The automaton, known as "Brian's Brain", runs for exactly 64 steps.
# The rules are applied simultaneously to every cell for each step:
# 1. An "Off" cell (0) turns "On" (2) if it has exactly two "On" neighbors.
# 2. An "On" cell (2) becomes "Firing" (1).
# 3. A "Firing" cell (1) turns "Off" (0).
for _ in range(64):
# The next state is calculated based on the current state.
new_grid = [[0] * GRID_SIZE for _ in range(GRID_SIZE)]
for r in range(GRID_SIZE):
for c in range(GRID_SIZE):
# Count "On" neighbors (state 2) in the current grid.
on_neighbors = 0
for dr in [-1, 0, 1]:
for dc in [-1, 0, 1]:
if dr == 0 and dc == 0:
continue
# The grid wraps around.
nr, nc = (r + dr) % GRID_SIZE, (c + dc) % GRID_SIZE
if grid[nr][nc] == 2:
on_neighbors += 1
# Apply rules based on the original grid state.
current_state = grid[r][c]
if current_state == 0 and on_neighbors == 2:
new_grid[r][c] = 2
elif current_state == 2:
new_grid[r][c] = 1
elif current_state == 1:
new_grid[r][c] = 0
else:
new_grid[r][c] = grid[r][c]
grid = new_grid
# The final state of the grid is converted to a string and hashed.
final_state_string = "".join(str(cell) for row in grid for cell in row)
final_hash = hashlib.sha256(final_state_string.encode('utf-8')).hexdigest()
# To solve the puzzle, the hash of the final state must match this target value.
return final_hash == "00415d5e21e055902d2039750b37494a50b8b67119e07572719541a318f77348"
import hashlib
def mystery(x):
"""
Accepts a string input and performs several independent, well-known
cryptographic and structural checks. Finding an input that satisfies
all of them is the goal.
"""
# Only accept strings.
if not isinstance(x, str):
return False
# Structural constraint: exact length (a subtle but strong filter).
if len(x) != 43:
return False
b = x.encode('utf-8')
# Multiple independent hash checks (these three together are highly
# constraining and correlate to a single well-known phrase).
if hashlib.md5(b).hexdigest() != "9e107d9d372bb6826bd81d3542a419d6":
return False
if hashlib.sha1(b).hexdigest() != "2fd4e1c67a2d28fced849ee1bb76e7391b93eb12":
return False
if hashlib.sha256(b).hexdigest() != (
"d7a8fbb307d7809469ca9abcb0082e4f"
"8d5651e46d3cdb762d02d0bf37c9e592"
):
return False
# A lightweight additional check: the ASCII sum of bytes modulo a prime.
# (This is redundant given the three hashes, but adds obfuscation.)
if sum(b) % 997 != 43:
return False
return True
import hashlib
import struct
def mystery(x):
"""
This puzzle requires finding a specific input string that, when hashed,
produces an initial state for a complex iterative system. The system
is then evolved for many steps, and the final state must match a
hidden target configuration.
"""
if not isinstance(x, str):
return False
# A small constraint to guide the search and prevent overly long inputs.
if len(x) > 64:
return False
try:
# Step 1: The input string is converted into a 256-bit seed using SHA-256.
# This makes the relationship between the input and the system's state opaque.
b = x.encode('utf-8')
h = hashlib.sha256(b).digest()
# Step 2: The 256-bit seed is used to initialize an 8-element state vector
# of 32-bit unsigned integers. The byte order is fixed but not obvious.
state = list(struct.unpack('<8I', h))
# Step 3: The state is evolved for a large number of rounds using a custom,
# non-linear mixing function. This process is computationally irreducible;
# there is no known shortcut to determine the final state from the initial one.
ROUNDS = 65536
def _mix_state(s):
# This function uses a combination of bitwise operations, arithmetic, and
# constant additions to thoroughly scramble the state in each round.
s[0] = (s[0] + s[1] + 0x9e3779b9) & 0xFFFFFFFF
s[2] = (s[2] ^ s[3]) & 0xFFFFFFFF
s[1] = ((s[1] << 5) | (s[1] >> 27)) & 0xFFFFFFFF # ROL 5
s[3] = (s[3] + s[4]) & 0xFFFFFFFF
s[4] = (s[4] ^ 0xdeadbeef) & 0xFFFFFFFF
s[5] = (s[5] + s[6] * 3) & 0xFFFFFFFF
s[7] = ((s[7] >> 11) | (s[7] << 21)) & 0xFFFFFFFF # ROR 11
s[6] = (s[6] ^ s[0]) & 0xFFFFFFFF
# The state elements are permuted to ensure mixing across the entire vector.
return [s[7], s[0], s[1], s[2], s[3], s[4], s[5], s[6]]
for _ in range(ROUNDS):
state = _mix_state(state)
# Step 4: The final state of the system after all rounds must exactly
# match this hardcoded target vector.
target_state = (
2994473347, 334863390, 169128531, 3881438906,
2690070267, 3215286280, 2407519630, 239276964
)
return tuple(state) == target_state
except Exception:
# Gracefully handle any errors during hashing or unpacking.
return False