def mystery(x):
if not isinstance(x, int):
return False
if x < 100 or x > 999:
return False
def is_prime(n):
if n < 2:
return False
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True
if not is_prime(x):
return False
digit_sum = sum(int(d) for d in str(x))
if not is_prime(digit_sum):
return False
reversed_x = int(str(x)[::-1])
if not is_prime(reversed_x):
return False
return True
import hashlib
def mystery(x):
# The input must be a string representation of a number.
if not isinstance(x, str) or not x.isdecimal():
return False
# The number must be of a certain magnitude.
if not (5 <= len(x) <= 8):
return False
val = int(x)
# A lower bound to narrow the search space slightly.
if val < 65536: # 2**16
return False
# The first major condition: a relationship between the number and its own hash.
# This requires finding a partial hash collision.
h = hashlib.sha256(x.encode()).hexdigest()
if not h.endswith(x[-4:]):
return False
# The second major condition: an obscure property of the number's binary representation.
# The numbers in this set are not obviously related without inspection.
magic_set = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}
if bin(val).count('1') not in magic_set:
return False
return True
def mystery(x):
if not isinstance(x, str):
return False
if len(x) < 5 or len(x) > 10:
return False
# Must contain only lowercase letters and digits
if not all(c.islower() or c.isdigit() for c in x):
return False
# Condition 1: Palindrome
if x != x[::-1]:
return False
# Condition 2: Contains at least one digit and at least one letter
if not any(c.isdigit() for c in x):
return False
if not any(c.isalpha() for c in x):
return False
# Condition 3: Sum of digit values equals number of letter characters
digit_sum = sum(int(c) for c in x if c.isdigit())
letter_count = sum(1 for c in x if c.isalpha())
if digit_sum != letter_count:
return False
# Condition 4: The string must contain at least 2 distinct letters
distinct_letters = len(set(c for c in x if c.isalpha()))
if distinct_letters < 2:
return False
return True
import struct
import hashlib
import math
def mystery(x):
# The input must be a string that can be parsed as a floating point number.
try:
val = float(x)
except (ValueError, TypeError):
return False
# The value must lie within a specific, narrow range.
if not (0.1 < val < 0.2):
return False
# Condition 1: An arcane property related to the number's binary representation.
# We consider the 64-bit IEEE 754 representation of the float.
try:
packed = struct.pack('!d', val)
int_rep = struct.unpack('!Q', packed)[0]
except (struct.error, OverflowError):
# This might happen for special values like NaN or infinity.
return False
# This integer representation must be a perfect square.
sqrt_int_rep = math.isqrt(int_rep)
if sqrt_int_rep * sqrt_int_rep != int_rep:
return False
# Condition 2: A cryptographic link between the original string and its binary form.
h = hashlib.md5(x.encode()).hexdigest()
# The number of set bits ('1's) in the binary integer representation is significant.
popcount = bin(int_rep).count('1')
# A substring of the hash must match a property of the popcount.
if hex(popcount)[2:] not in h:
return False
return True
def mystery(x):
if not isinstance(x, int):
return False
if x < 1000 or x > 9999:
return False
s = str(x)
# Condition 1: all digits must be non-zero and even
for d in s:
if int(d) == 0 or int(d) % 2 != 0:
return False
# Condition 2: product of first two digits equals product of last two digits
prod_first = int(s[0]) * int(s[1])
prod_last = int(s[2]) * int(s[3])
if prod_first != prod_last:
return False
# Condition 3: sum of all digits must be divisible by 6
digit_sum = sum(int(d) for d in s)
if digit_sum % 6 != 0:
return False
# Condition 4: the number must be divisible by 11
if x % 11 != 0:
return False
return True
import math
def mystery(x):
# The input must be an integer.
if not isinstance(x, int):
return False
# Condition 1: The number must be within a specific range.
if not (1_000_000 < x < 100_000_000):
return False
# Condition 2: The number must be a Fibonacci number.
# This is tested using the property that for a number n > 0,
# either 5*n^2 + 4 or 5*n^2 - 4 must be a perfect square.
def is_perfect_square(n):
if n < 0: return False
if n == 0: return True
s = math.isqrt(n)
return s * s == n
if not (is_perfect_square(5 * x * x + 4) or is_perfect_square(5 * x * x - 4)):
return False
# Condition 3: The number must be a semiprime.
# A semiprime is a product of exactly two prime numbers (p*q), not necessarily distinct.
# To check this, we find the smallest prime factor, p. Then we check if x/p is also prime.
limit = math.isqrt(x)
p = 0
# Find the smallest factor p. Since it's the smallest, it must be prime.
if x % 2 == 0:
p = 2
else:
# Check odd factors up to sqrt(x)
for i in range(3, limit + 1, 2):
if x % i == 0:
p = i
break
# If no factor is found up to sqrt(x), then x itself is prime.
# A prime number is not a semiprime, so we return False.
if p == 0:
return False
q = x // p
# If x = p*p*p..., the smallest factor p is found, but q = x/p is not prime.
# We must confirm q is prime.
# Primality test for q
if q < 2:
return False
if q == 2:
# q is prime. The check is complete.
return True
if q % 2 == 0:
# q is not prime (and not 2)
return False
q_limit = math.isqrt(q)
for i in range(3, q_limit + 1, 2):
if q % i == 0:
# q has a factor, so it's not prime.
return False
# If we've reached here, q is prime.
# So x = p * q, where p is the smallest prime factor and q is also prime.
# This means x is a semiprime.
return True
def mystery(x):
if not isinstance(x, str):
return False
# Must be exactly 7 characters
if len(x) != 7:
return False
# Must be all lowercase letters
if not x.islower() or not x.isalpha():
return False
# Condition 1: Must contain exactly 3 distinct letters
if len(set(x)) != 3:
return False
# Condition 2: Each distinct letter must appear at least twice
from collections import Counter
counts = Counter(x)
if any(count < 2 for count in counts.values()):
return False
# Condition 3: First and last characters must be the same
if x[0] != x[-1]:
return False
# Condition 4: The string must be a palindrome
if x != x[::-1]:
return False
return True
def mystery(x):
# --- Input Validation ---
if not isinstance(x, str):
return False
if not all(c in 'NESW' for c in x):
return False
# Path length constraint. This is the length of the shortest path.
if len(x) != 48:
return False
# --- Checksum Constraint ---
# The sum of ASCII values of the path characters must match a target.
# This uniquely identifies the correct path among multiple shortest paths.
checksum = sum(ord(c) for c in x)
if checksum != 3644:
return False
# --- Path Traversal Simulation ---
r, c = 0, 0
start_pos = (0, 0)
end_pos = (15, 15)
# The maze's wall generation logic.
def _is_wall(r1, c1, r2, c2):
# Check for out-of-bounds moves.
if r1 < 0 or r1 > 15 or c1 < 0 or c1 > 15 or \
r2 < 0 or r2 > 15 or c2 < 0 or c2 > 15:
return True
# Ensure symmetry by ordering coordinates.
if r1 > r2: r1, r2 = r2, r1
if c1 > c2: c1, c2 = c2, c1
# A pseudo-random but deterministic function based on coordinates.
if r1 == r2: # Horizontal wall check
seed = r1 * 101 + c1 * 103
else: # Vertical wall check
seed = r1 * 107 + c1 * 113
# Use a large prime multiplier for better pseudo-randomness.
val = (seed * 2654435761) & 0xFFFFFFFF
# A 35% chance of a wall between any two adjacent cells.
return (val % 100) < 35
for move in x:
nr, nc = r, c
if move == 'N': nr -= 1
elif move == 'S': nr += 1
elif move == 'W': nc -= 1
elif move == 'E': nc += 1
if _is_wall(r, c, nr, nc):
return False # The path is invalid if it hits a wall.
r, c = nr, nc
# --- Goal Check ---
# The path must start at (0,0) and end at the target location (15, 15).
return (r, c) == end_pos
def mystery(x):
if not isinstance(x, str):
return False
if len(x) != 8:
return False
# Must contain only lowercase letters
if not x.islower() or not x.isalpha():
return False
# Condition 1: Must contain exactly 4 distinct letters
if len(set(x)) != 4:
return False
# Condition 2: Each distinct letter must appear exactly twice
from collections import Counter
counts = Counter(x)
if any(count != 2 for count in counts.values()):
return False
# Condition 3: No two adjacent characters can be the same
for i in range(len(x) - 1):
if x[i] == x[i+1]:
return False
# Condition 4: The letters at even positions (0,2,4,6) must all be different
if len(set([x[0], x[2], x[4], x[6]])) != 4:
return False
# Condition 5: First letter must come before last letter alphabetically
if x[0] >= x[-1]:
return False
return True
import math
def mystery(x):
# --- Input Validation ---
if not isinstance(x, str) or len(x) != 20:
return False
if not all(c in '01' for c in x):
return False
# --- Condition 1: A Number Theoretic Property ---
# The integer representation of the binary string must be a prime number.
try:
val = int(x, 2)
except ValueError:
return False
def is_prime(n):
if n < 2: return False
if n == 2: return True
if n % 2 == 0: return False
limit = math.isqrt(n)
for i in range(3, limit + 1, 2):
if n % i == 0:
return False
return True
if not is_prime(val):
return False
# --- Condition 2: An Iterative Transformation ---
# The input string is treated as an initial state that evolves over time.
# It must evolve into a specific target state after a fixed number of steps.
# The update rule for the transformation.
def _update_rule(l, c, r):
# (left OR center) XOR right
return (l | c) ^ r
num_steps = 10
length = 20
state = [int(c) for c in x]
for _ in range(num_steps):
next_state = [0] * length
for i in range(length):
# The state array wraps around (periodic boundary conditions)
left = state[(i - 1 + length) % length]
center = state[i]
right = state[(i + 1) + length) % length]
next_state[i] = _update_rule(left, center, right)
state = next_state
final_state_str = "".join(map(str, state))
target_state = '10101010101010101010'
return final_state_str == target_state