def mystery(x):
if not isinstance(x, str) or len(x) != 4:
return False
v = [ord(c) for c in x]
return (
((v[0] << 2) ^ v[1] == 0x16C) and
((v[1] ^ v[2]) == 0x47) and
(v[0] + v[3] == 0x78) and
((v[2] * v[3]) % 100 == 0x1B)
)
import struct
import math
def mystery(x):
"""
This function takes a value x and returns True only for a very specific
64-bit float, determined by three obscure properties of its underlying
binary representation.
"""
# The puzzle requires a specific 64-bit float.
if not isinstance(x, float):
return False
# Avoid non-finite numbers which have special representations.
if not math.isfinite(x):
return False
try:
# Convert the float to its 64-bit big-endian integer representation.
# This is the core of the puzzle: understanding IEEE 754.
bits = struct.unpack('>Q', struct.pack('>d', x))[0]
# First condition: A checksum-like operation on the high 32 bits.
# This part is difficult to reverse-engineer.
p1 = (bits >> 32)
if (p1 * 13 + 7) % 1000000 != 592585:
return False
# Second condition: A bitwise operation mixing high and low parts.
# This obfuscates the relationship between the parts of the float.
p2 = (bits & 0xFFFFFFFF)
if ((p1 >> 8) ^ p2) != 0x8b5452d6:
return False
# Third condition: A property of the entire bit pattern.
# The number of set bits (population count) must be 32.
# This is another non-obvious property that is hard to work back from.
if bin(bits).count('1') != 32:
return False
# If all three obscure conditions are met, we have found the number.
return True
except (struct.error, OverflowError):
# Catch potential errors with weird float values.
return False
import struct
def mystery(x):
# The input must be an 8-character string.
if not isinstance(x, str) or len(x) != 8:
return False
# Restrict to standard ASCII to ensure consistent byte encoding.
if any(ord(c) > 127 for c in x):
return False
# Convert the string into two 32-bit big-endian integers.
# This splits the 8 bytes into a left part (u) and a right part (v).
u, v = struct.unpack('>II', x.encode('ascii'))
# Perform 2 rounds of a Feistel-like network.
# This mixes the two halves reversibly using a non-linear function.
for _ in range(2):
# The mixing function involves addition with a constant and masking.
# Constant: 0xDEADBEEF (3735928559)
f = (v + 0xDEADBEEF) & 0xFFFFFFFF
# Standard Feistel swap and XOR:
# The new left becomes the old right.
# The new right becomes the old left XOR the function of the old right.
u, v = v, u ^ f
# The transformed integers must match these specific target values.
return u == 0x0C40837B and v == 0x8F9C634B
import hashlib
import re
def mystery(x):
"""
This function subjects the input x to a series of tests from
different domains: classical numeration, number theory, and cryptography.
Only a very specific string will pass all checks.
"""
# Helper function to test for primality using an optimized trial division method.
def is_prime(n):
if not isinstance(n, int) or n < 2:
return False
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
return True
# Helper to convert a Roman numeral string to an integer.
# It uses a regular expression to strictly validate the standard format.
def roman_to_int(s):
pattern = r'^M{0,3}(CM|CD|D?C{0,3})(XC|XL|L?X{0,3})(IX|IV|V?I{0,3})$'
if not isinstance(s, str) or not re.match(pattern, s) or not s:
return None # Invalid format or type
rom_val = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
int_val = 0
for i in range(len(s)):
if i > 0 and rom_val[s[i]] > rom_val[s[i-1]]:
# This handles subtractive cases like IV, IX, XL, etc.
int_val += rom_val[s[i]] - 2 * rom_val[s[i-1]]
else:
int_val += rom_val[s[i]]
return int_val
# --- Main puzzle logic ---
# The input must be a string composed of valid Roman numeral characters.
if not isinstance(x, str) or not all(c in 'IVXLCDM' for c in x):
return False
# Condition 1: The string must represent a valid Roman numeral.
# The helper function will return None if the format is invalid.
n = roman_to_int(x)
if n is None or n >= 4000: # We cap the number to keep the search space reasonable.
return False
# Condition 2: The integer value of the Roman numeral must be a prime number.
if not is_prime(n):
return False
# Condition 3: The SHA-1 hash of the lowercase string must end with a specific suffix.
# This is computationally irreversible and acts as a strong filter.
h = hashlib.sha1(x.lower().encode()).hexdigest()
if not h.endswith('f44'):
return False
# Condition 4: The string, when interpreted as a base-36 number (A=10, ..., Z=35),
# must satisfy a specific modular congruence. This is another non-obvious mathematical constraint.
try:
m = int(x, 36)
if m % 9973 != 2730:
return False
except ValueError:
return False
# If a value passes all these diverse and unrelated checks, it is the solution.
return True
def mystery(x):
# The input must be a 6-character string consisting of lowercase English letters.
if not isinstance(x, str) or len(x) != 6:
return False
if not all('a' <= c <= 'z' for c in x):
return False
# Map characters 'a'-'z' to integers 0-25.
v = [ord(c) - 97 for c in x]
# Constraint 1: The sum of the values must be 100.
if sum(v) != 100:
return False
# Constraint 2: The weighted sum (1-based index) must be 309.
# Formula: 1*v[0] + 2*v[1] + ... + 6*v[5]
if sum((i + 1) * val for i, val in enumerate(v)) != 309:
return False
# Constraint 3: The alternating sum must be 2.
# Formula: v[0] - v[1] + v[2] - v[3] + v[4] - v[5]
if v[0] - v[1] + v[2] - v[3] + v[4] - v[5] != 2:
return False
# Constraint 4: The product of the first three values modulo 101 must be 26.
if (v[0] * v[1] * v[2]) % 101 != 26:
return False
# Constraint 5: The XOR sum of the last three values must be 22.
if v[3] ^ v[4] ^ v[5] != 22:
return False
# Constraint 6: The sum of squares of the first and last values must be 241.
if v[0]**2 + v[5]**2 != 241:
return False
return True
import binascii
# The user will need to install pycryptodome: pip install pycryptodome
from Crypto.Cipher import AES
def mystery(x):
"""
This function tests an input x against three distinct types of constraints:
structural, arithmetic, and cryptographic. Only one specific value of x
will satisfy all three.
"""
# Constraint 1: Input must be a 32-character hex string.
if not isinstance(x, str) or len(x) != 32:
return False
try:
# Ensure all characters are valid hex digits.
int(x, 16)
except ValueError:
return False
# Constraint 2: The string must be a palindrome.
# This is a major structural constraint that drastically reduces the
# effective search space, but still leaves it impossibly large
# without more information.
if x != x[::-1]:
return False
# Constraint 3: An arithmetic property. The sum of the integer values
# of the hexadecimal digits (0-15) must be exactly 180. This works
# in tandem with the palindrome constraint to define a specific,
# non-trivial subset of keys that a solver must generate and test.
if sum(int(c, 16) for c in x) != 180:
return False
# The final lock: The input string, when interpreted as a 128-bit key,
# must correctly decrypt a given ciphertext. This is computationally
# irreversible and serves as the ultimate validator.
try:
key = binascii.unhexlify(x)
# This is the ciphertext, encrypted with the one true key.
ciphertext = binascii.unhexlify(
"c32087520b88d3289053b26338b2552e"
"148286a1bf444b2f4c9a888c3086a343"
)
# The known plaintext that results from correct decryption.
expected_plaintext = b"The secret lies in dual constraints."
cipher = AES.new(key, AES.MODE_ECB)
decrypted = cipher.decrypt(ciphertext)
# If the decryption is successful, we have found the solution.
return decrypted == expected_plaintext
except (ValueError, binascii.Error):
# Handle potential errors with hex decoding.
return False
import struct
def mystery(x):
# The input must be an 8-character ASCII string.
if not isinstance(x, str) or len(x) != 8:
return False
# Ensure characters are standard ASCII.
try:
b = x.encode('ascii')
except UnicodeEncodeError:
return False
# Interpret the string as two 32-bit big-endian unsigned integers.
u, v = struct.unpack('>II', b)
# Operation 1: Add v to u (modulo 2^32)
u = (u + v) & 0xFFFFFFFF
# Operation 2: XOR v with the new u
v = v ^ u
# Operation 3: Rotate u left by 13 bits
# We simulate 32-bit rotation using shifts and a mask.
u = ((u << 13) | (u >> 19)) & 0xFFFFFFFF
# Operation 4: Add the rotated u to v (modulo 2^32)
v = (v + u) & 0xFFFFFFFF
# The transformed values must match these specific targets.
return u == 0x53736bcd and v == 0x24332687
def mystery(x):
"""
This function interprets the input string x as a program for a tiny virtual
machine. The program must drive the VM to a specific target state.
"""
# Constraint 1: Input must be a 22-character string.
if not isinstance(x, str) or len(x) != 22:
return False
# Constraint 2: The program must only use characters from the valid instruction set.
valid_chars = '><+-abc@?'
if not all(c in valid_chars for c in x):
return False
# --- VM Initialization ---
# A pre-initialized 16-byte memory tape.
mem = [15, 8, 100, 7, 21, 0, 12, 44, 9, 81, 3, 25, 60, 18, 5, 33]
mem_len = len(mem)
# Three 8-bit registers that wrap on overflow.
a, b, c = 0, 0, 0
# A pointer to the current position on the memory tape.
ptr = 0
# A program counter to iterate through the instructions.
pc = 0
# --- VM Execution Loop ---
# The VM runs for exactly the length of the program.
while pc < len(x):
instr = x[pc]
pc += 1
if instr == '>': # Move pointer right
ptr = (ptr + 1) % mem_len
elif instr == '<': # Move pointer left
ptr = (ptr - 1 + mem_len) % mem_len
elif instr == '+': # Increment memory value at pointer
mem[ptr] = (mem[ptr] + 1) % 256
elif instr == '-': # Decrement memory value at pointer
mem[ptr] = (mem[ptr] - 1 + 256) % 256
elif instr == 'a': # Add memory value to register 'a'
a = (a + mem[ptr]) % 256
elif instr == 'b': # Add memory value to register 'b'
b = (b + mem[ptr]) % 256
elif instr == 'c': # Add memory value to register 'c'
c = (c + mem[ptr]) % 256
elif instr == '@':
# Conditional skip: if register 'a' is even, skip the next instruction.
if a % 2 == 0:
pc += 1
elif instr == '?':
# Conditional skip: if memory at ptr is zero, skip the next instruction.
if mem[ptr] == 0:
pc += 1
# --- Final State Check ---
# The program is only "correct" if it leaves the registers in this exact state.
return a == 177 and b == 88 and c == 201
import struct
def mystery(x):
# Constraint 1: Input must be a 12-character string.
if not isinstance(x, str) or len(x) != 12:
return False
# Constraint 2: All characters must be printable ASCII (32-126).
# This ensures the solution is a "clean" string.
if not all(32 <= ord(c) <= 126 for c in x):
return False
try:
# Convert the string into three 32-bit big-endian integers.
# v[0] corresponds to chars 0-3
# v[1] corresponds to chars 4-7
# v[2] corresponds to chars 8-11
v = struct.unpack('>III', x.encode('ascii'))
except Exception:
return False
# The integers must satisfy a coupled system of equations involving
# XOR, Modular Addition, and Bitwise Rotation.
# Equation 1: XOR relationship between first and second integers.
if (v[0] ^ v[1]) != 0x21261506:
return False
# Equation 2: Additive relationship between second and third integers.
if (v[1] + v[2]) & 0xFFFFFFFF != 0xE4B8D984:
return False
# Equation 3: A mix of Rotation and XOR involving the third and first integers.
# Rotate v[2] right by 11 bits (circular shift).
rot_v2 = ((v[2] >> 11) | (v[2] << 21)) & 0xFFFFFFFF
if (rot_v2 ^ v[0]) != 0xD75B3CCB:
return False
return True
import hashlib
def mystery(x):
"""
This function treats the input string as the initial state for a
cellular automaton and checks if its evolution matches a hidden pattern.
"""
# Constraint: Input must be an 8-character ASCII string.
if not isinstance(x, str) or len(x) != 8:
return False
try:
b = x.encode('ascii')
except UnicodeEncodeError:
return False
# Convert the 8 bytes (64 bits) into the initial state (generation 0).
state = []
for byte in b:
for i in range(8):
state.append((byte >> (7 - i)) & 1)
width = len(state)
num_generations = 32
# Evolve the state for 32 generations using Wolfram's Rule 30.
# The boundary conditions are periodic (the grid wraps around).
for _ in range(num_generations):
next_state = [0] * width
for i in range(width):
# Get the 3-cell neighborhood: left, center, right.
left = state[(i - 1 + width) % width]
center = state[i]
right = state[(i + 1) % width]
# Apply Rule 30: new_center = left XOR (center OR right)
next_state[i] = left ^ (center | right)
state = next_state
# The final state must produce a specific SHA256 hash.
# Convert the final bit array back to bytes for hashing.
final_bytes = bytearray()
for i in range(0, width, 8):
byte = 0
for j in range(8):
byte = (byte << 1) | state[i + j]
final_bytes.append(byte)
h = hashlib.sha256(final_bytes).hexdigest()
return h == "c87eb9548425c276189b58117283622f615363c34a2e519280d44339f408992a"