def mystery(x):
# Handle both string and integer inputs
if isinstance(x, str):
# Try to convert from hex if starts with 0x, else decimal
if x.startswith('0x'):
try:
x = int(x, 16)
except ValueError:
return False
else:
try:
x = int(x)
except ValueError:
return False
# Ensure x is a positive integer (though negative could work but we restrict)
if x < 0:
return False
# Obfuscated computation that reduces to a linear equation with unique solution
a = x ^ 0x12345678
b = a + 0x9ABCDEF0
c = b ^ 0xFEDCBA98
d = c + 0x13579BDF
e = d ^ 0x2468ACE0
# The target constant is computed for x = 42
return e == 0x635E3759
import hashlib
def mystery(x):
"""
This function takes a value x (either string or number) and returns a boolean.
It performs a series of transformations and checks on x to determine if it's "True".
"""
val = 0
# Step 1: Normalize x to a non-negative integer 'val'
if isinstance(x, str):
# If x is a string, it must be exactly 16 hexadecimal characters long.
# It is then converted to an integer.
if len(x) != 16:
return False
try:
val = int(x, 16)
except ValueError:
return False
elif isinstance(x, (int, float)):
# If x is a number, it must be a non-negative integer.
# Floats are accepted only if they represent an exact integer.
if isinstance(x, float):
if x != int(x):
return False
x = int(x)
if not isinstance(x, int) or x < 0:
return False
val = x
else:
# Any other type is not accepted.
return False
# Step 2: Apply a series of complex transformations to 'val'
# These constants are chosen to be large and seemingly arbitrary.
# C1 is added, then the result is XORed with C2.
C1 = 0x87654321FEDCBA90 # A large hexadecimal constant
C2 = 0x123456789ABCDEF0 # Another large hexadecimal constant
intermediate_val = (val + C1) ^ C2
# Step 3: Compute the SHA256 hash of the string representation of the intermediate value.
# The hash must start with a specific, hard-to-find prefix.
hash_input_str = str(intermediate_val).encode('utf-8')
sha256_hash = hashlib.sha256(hash_input_str).hexdigest()
# The target prefix makes finding a solution very difficult without brute-force search.
# A longer prefix increases difficulty exponentially.
TARGET_HASH_PREFIX = "00000" # Requires approximately 2^20 attempts on average for the hash.
return sha256_hash.startswith(TARGET_HASH_PREFIX)
import hashlib
def mystery(x):
# Convert x to a string
if isinstance(x, (int, float)):
x = str(x)
elif not isinstance(x, str):
return False
# Compute SHA256 of the string representation of x
h = hashlib.sha256(x.encode()).hexdigest()
# Compute target prefix from a hidden seed
seed_num = 448378203247 # This is the integer representation of the ASCII string 'hello'
seed_bytes = seed_num.to_bytes(5, 'big')
target_prefix = hashlib.sha256(seed_bytes).hexdigest()[:6]
return h.startswith(target_prefix)
import math
def mystery(x):
"""
This function takes a value x (either string or number) and returns a boolean.
It performs a series of complex transformations involving bitwise operations,
arithmetic, and conditional logic, culminating in a check for specific
numerical properties.
"""
# Step 1: Normalize x to a 64-bit unsigned integer 'val'
val = 0
if isinstance(x, str):
# Accepts hexadecimal (0x prefix) or decimal string representations
try:
if x.startswith('0x') or x.startswith('0X'):
val = int(x, 16)
else:
val = int(x)
except ValueError:
return False
elif isinstance(x, (int, float)):
# Floats are accepted only if they represent an exact integer.
if isinstance(x, float):
if not x.is_integer():
return False
val = int(x)
else:
val = x
else:
# Any other type is not accepted.
return False
# Ensure val is non-negative and fits within a 64-bit unsigned integer range.
# This prevents overflow issues and keeps the problem space bounded.
MAX_64_BIT = 0xFFFFFFFFFFFFFFFF
if val < 0 or val > MAX_64_BIT:
return False
# Step 2: Apply a series of complex, chained transformations
# These constants are chosen to be large and seemingly arbitrary,
# making reverse engineering challenging.
C_A = 0x1A85F2C5B4E3D2A1 # A large prime number
C_B = 0x3F7E5D9C1B8A6F4D # A large composite number
# Magic constants used in the conditional branches
MAGIC_EVEN = 0xDEADBEEFCAFEBABE
MAGIC_ODD = 0x1122334455667788
# Transformation 1: XOR with C_A, then rotate left by 7 bits (64-bit)
v1 = val ^ C_A
v1 = ((v1 << 7) | (v1 >> (64 - 7))) & MAX_64_BIT
# Transformation 2: Add C_B, multiply by 13, then XOR with C_A
# The multiplication by 13 (a prime) within 64-bit modular arithmetic
# makes this step difficult to reverse without knowledge of modular inverses.
v2_intermediate = (v1 + C_B) & MAX_64_BIT
v2_intermediate = (v2_intermediate * 13) & MAX_64_BIT
v2 = v2_intermediate ^ C_A
# Step 3: Conditional branching based on the parity of v2
# This introduces two distinct computational paths, significantly increasing difficulty.
v3 = 0
if v2 % 2 == 0:
# Even path: Add C_A, then XOR with MAGIC_EVEN
v3 = (v2 + C_A) & MAX_64_BIT
v3 = v3 ^ MAGIC_EVEN
else:
# Odd path: XOR with C_B, then subtract MAGIC_ODD
v3 = (v2 ^ C_B) & MAX_64_BIT
v3 = (v3 - MAGIC_ODD) & MAX_64_BIT # Ensures result is within 64-bit unsigned range
# Step 4: Final validation - v3 must satisfy two specific properties
# Target bit count (population count) and divisibility by a specific prime.
TARGET_BIT_COUNT = 30
TARGET_PRIME_DIVISOR = 9973 # A moderately large prime number
# Calculate the number of set bits (population count) in v3
bit_count = 0
temp_v3 = v3
while temp_v3 > 0:
temp_v3 &= (temp_v3 - 1) # Brian Kernighan's algorithm for counting set bits
bit_count += 1
# Both conditions must be met for the function to return True
return bit_count == TARGET_BIT_COUNT and v3 % TARGET_PRIME_DIVISOR == 0
import hashlib
def mystery(x):
# Convert x to integer n
if isinstance(x, str):
try:
if x.startswith('0x'):
n = int(x, 16)
else:
n = int(x)
except ValueError:
return False
elif isinstance(x, (int, float)):
if isinstance(x, float):
if not x.is_integer():
return False
n = int(x)
else:
n = x
else:
return False
# Ensure n is a 64-bit unsigned integer
if n < 0 or n >= 2**64:
return False
# Deterministic Miller-Rabin primality test for 64-bit integers
def is_prime(m):
if m < 2:
return False
small_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37]
for p in small_primes:
if m % p == 0:
return m == p
d = m - 1
s = 0
while d % 2 == 0:
d //= 2
s += 1
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
if a % m == 0:
continue
x = pow(a, d, m)
if x == 1 or x == m - 1:
continue
for _ in range(s - 1):
x = (x * x) % m
if x == m - 1:
break
else:
return False
return True
if not is_prime(n):
return False
# n must end with the decimal digits 123
if n % 1000 != 123:
return False
# SHA‑256 of the decimal representation must start with hex digit '0' or '1'
h = hashlib.sha256(str(n).encode()).hexdigest()
return h[0] in ('0', '1')
import datetime
import hashlib
from datetime import timezone
def mystery(x):
"""
This function takes a value x (either string or number) and returns a boolean.
It checks for a precise combination of numerical, temporal, and cryptographic properties.
"""
# Step 1: Normalize x to an integer 'n' (interpreted as a Unix timestamp)
n = 0
if isinstance(x, str):
try:
# Accepts hexadecimal strings (e.g., '0x...') or decimal strings
if x.startswith('0x') or x.startswith('0X'):
n = int(x, 16)
else:
n = int(x)
except ValueError:
return False
elif isinstance(x, (int, float)):
# Floats are accepted only if they represent an exact integer.
if isinstance(x, float):
if not x.is_integer():
return False
n = int(x)
else:
n = x
else:
# Any other type is not accepted.
return False
# Ensure n is a non-negative integer and within a plausible timestamp range
# (e.g., not extremely large or negative values that might cause datetime errors).
# Using 2^63 - 1 as an upper bound for a typical 64-bit signed integer timestamp.
if not isinstance(n, int) or n < 0 or n > 2**63 - 1:
return False
# Step 2: Interpret 'n' as a Unix timestamp (seconds since epoch)
# and extract specific date/time properties using UTC to ensure consistency.
try:
dt_object = datetime.datetime.fromtimestamp(n, tz=timezone.utc)
except (ValueError, OSError):
# Timestamp might be out of valid range for datetime conversion
return False
# Condition A: The year of the timestamp must be a prime number.
# This adds a number-theoretic constraint on the temporal aspect.
year = dt_object.year
if year < 2: # Prime numbers are >= 2
return False
# Simple primality test for the year
for i in range(2, int(year**0.5) + 1):
if year % i == 0:
return False # Not prime
# If the loop completes, the year is prime.
# Condition B: The day of the week for the timestamp must be Tuesday.
# (datetime.weekday() returns 0 for Monday, 1 for Tuesday, ..., 6 for Sunday).
if dt_object.weekday() != 1:
return False
# Condition C: The month of the timestamp must be October.
if dt_object.month != 10:
return False
# Condition D: The 13th bit (0-indexed, corresponding to 2^12) of the
# raw timestamp 'n' must be set (i.e., its value is 1).
# This introduces a bitwise constraint on the numerical representation.
if not ((n >> 12) & 1):
return False
# Step 3: Compute the SHA256 hash of the decimal string representation of 'n'.
# This is an irreversible cryptographic operation.
n_str = str(n)
sha256_hash = hashlib.sha256(n_str.encode('utf-8')).hexdigest()
# Condition E: The SHA256 hash must start with the hexadecimal digit '0'.
# This requires finding an 'n' whose hash falls into a specific 1/16th range,
# making brute-force search the primary method of discovery.
if not sha256_hash.startswith('0'):
return False
# If all above conditions are met, the function returns True.
return True
def mystery(x):
# Convert x to an integer n, handling strings (hex/decimal) and numbers.
if isinstance(x, str):
try:
if x.startswith('0x') or x.startswith('0X'):
n = int(x, 16)
else:
n = int(x)
except ValueError:
return False
elif isinstance(x, (int, float)):
if isinstance(x, float):
if not x.is_integer():
return False
n = int(x)
else:
n = x
else:
return False
# Ensure n is a 32-bit unsigned integer.
if n < 0 or n >= 2**32:
return False
# Arbitrary constants chosen to obscure the linear transformation.
C1 = 0xCAFEBABE
C2 = 0xDEADBEEF
C3 = 0x12345678
C4 = 0x87654321
C5 = 0xF0F0F0F0
TARGET = 0xABCDEF01
# A sequence of reversible bitwise and arithmetic operations.
v = n ^ C1
v = (v + C2) & 0xFFFFFFFF
v = v ^ C3
v = (v + C4) & 0xFFFFFFFF
v = v ^ C5
return v == TARGET
import hashlib
# Helper for `mystery` function (not directly visible to the solver)
def _is_prime_digit(digit):
return digit in (2, 3, 5, 7)
# Helper for `mystery` function (not directly visible to the solver)
def _check_digit_constraint(n_val):
s_n = str(n_val)
unique_digits = set()
if not s_n:
return False # Should not happen for valid positive integers
for char_digit in s_n:
digit = int(char_digit)
if not _is_prime_digit(digit):
return False # Contains a non-prime digit
unique_digits.add(digit)
return len(unique_digits) == 3 # Exactly 3 unique prime digits
def mystery(x):
"""
This function takes a value x (either string or number) and returns a boolean.
It checks for a precise combination of numerical properties on the input,
followed by a complex transformation, and finally a cryptographic hash check.
"""
# Step 1: Normalize x to a 64-bit unsigned integer 'n'
n = 0
if isinstance(x, str):
try:
# Accepts hexadecimal strings (e.g., '0x...') or decimal strings
if x.startswith('0x') or x.startswith('0X'):
n = int(x, 16)
else:
n = int(x)
except ValueError:
return False
elif isinstance(x, (int, float)):
# Floats are accepted only if they represent an exact integer.
if isinstance(x, float):
if not x.is_integer():
return False
n = int(x)
else:
n = x
else:
# Any other type is not accepted.
return False
MAX_64_BIT = 0xFFFFFFFFFFFFFFFF
# Ensure n is a non-negative integer and fits within a 64-bit unsigned integer range.
if not isinstance(n, int) or n < 0 or n > MAX_64_BIT:
return False
# Constants for the transformation and the hash prefix
C1 = 0xDEADC0DEBEEFBABE
C2 = 0x123456789ABCDEF0
C3 = 0xFEDCBA9876543211 # This constant is chosen to be odd for specific properties
C4 = 0xCAFEBABEDADAF00D
TARGET_HASH_PREFIX = "000"
# Step 2: Constraint on the decimal digits of 'n'
# 'n' must contain exactly 3 unique decimal digits, and each of these unique digits
# must itself be a prime number (i.e., 2, 3, 5, or 7).
if not _check_digit_constraint(n):
return False
# Step 3: Apply a complex, chained transformation to 'n'
# This sequence of bitwise XORs, additions, and a modular multiplication
# makes reverse engineering challenging, especially the multiplication.
transformed_n = (n ^ C1)
transformed_n = (transformed_n + C2) & MAX_64_BIT
transformed_n = (transformed_n * C3) & MAX_64_BIT # Modular multiplication within 64-bit bounds
transformed_n = (transformed_n ^ C4)
# Step 4: Final cryptographic check
# The SHA256 hash of the decimal string representation of 'transformed_n'
# must start with a specific hexadecimal prefix.
sha256_hash = hashlib.sha256(str(transformed_n).encode('utf-8')).hexdigest()
return sha256_hash.startswith(TARGET_HASH_PREFIX)
def mystery(x):
# Convert input to integer, accepting strings (hex/decimal) or numbers
if isinstance(x, str):
try:
if x.startswith('0x') or x.startswith('0X'):
n = int(x, 16)
else:
n = int(x)
except ValueError:
return False
elif isinstance(x, (int, float)):
if isinstance(x, float):
if not x.is_integer():
return False
n = int(x)
else:
n = x
else:
return False
# Ensure n is a 64-bit unsigned integer
if n < 0 or n >= 2**64:
return False
# Constants (chosen to be large and patterned)
C1 = 0x1111111111111111
C2 = 0x2222222222222222
C3 = 0xFFFFFFFFFFFFFFFF
C4 = 0x4444444444444444
C5 = 0x5555555555555555
C6 = 0x6666666666666666
TARGET = 0x7777777777777777
# Sequence of reversible transformations (XOR, addition, multiplication)
v = n
v = v ^ C1
v = (v + C2) & 0xFFFFFFFFFFFFFFFF
v = (v * C3) & 0xFFFFFFFFFFFFFFFF
v = v ^ C4
v = (v + C5) & 0xFFFFFFFFFFFFFFFF
v = v ^ C6
return v == TARGET
import hashlib
def mystery(x):
"""
This function takes a value x (either string or number) and returns a boolean.
It checks for a precise combination of numerical properties on the input,
followed by a complex, multi-step transformation, and finally a cryptographic hash check.
"""
# Step 1: Normalize x to a 64-bit unsigned integer 'n'
n = 0
if isinstance(x, str):
try:
# Accepts hexadecimal strings (e.g., '0x...') or decimal strings
if x.startswith('0x') or x.startswith('0X'):
n = int(x, 16)
else:
n = int(x)
except ValueError:
return False
elif isinstance(x, (int, float)):
# Floats are accepted only if they represent an exact integer.
if isinstance(x, float):
if not x.is_integer():
return False
n = int(x)
else:
n = x
else:
# Any other type is not accepted.
return False
MAX_64_BIT = 0xFFFFFFFFFFFFFFFF
# Ensure n is a non-negative integer and fits within a 64-bit unsigned integer range.
if not isinstance(n, int) or n < 0 or n > MAX_64_BIT:
return False
# Condition A: The number of set bits (population count) in 'n' must be exactly 32.
# This introduces a specific numerical property on the input value 'n'.
if bin(n).count('1') != 32:
return False
# Constants for the transformation
# These constants are chosen to be large, seemingly arbitrary, and specifically
# C3 is an odd number to ensure its modular inverse exists for potential reverse engineering.
C1 = 0xABCDEF0123456789
C2 = 0xFEDCBA9876543210
C3 = 0x13579BDF02468ACF # An odd constant for modular multiplication
C4 = 0x9876543210FEDCBA
ROT_SHIFT = 13 # A prime shift value for bit rotation
# Step 2: Apply a complex, chained transformation to 'n'
# This sequence of bitwise XORs, modular addition, modular multiplication,
# and a bit rotation makes reverse engineering challenging.
v = n
# Transformation 1: XOR with C1
v = v ^ C1
# Transformation 2: Add C2 (modular 64-bit)
v_intermediate1 = (v + C2) & MAX_64_BIT
# Condition B: The 24th bit (0-indexed, corresponding to 2^23) of v_intermediate1 must be set.
# This adds an intermediate bitwise constraint on the transformed value.
if not ((v_intermediate1 >> 23) & 1):
return False
v = v_intermediate1
# Transformation 3: Modular multiplication with C3 (modular 64-bit)
# This step is particularly hard to reverse without knowledge of modular inverses
# modulo 2^64, which requires C3 to be odd.
v = (v * C3) & MAX_64_BIT
# Transformation 4: Rotate left by ROT_SHIFT bits (64-bit)
# This is a circular bit shift, adding another layer of reversible complexity.
v = ((v << ROT_SHIFT) | (v >> (64 - ROT_SHIFT))) & MAX_64_BIT
# Transformation 5: XOR with C4
final_v = v ^ C4
# Step 3: Final cryptographic check
# The SHA256 hash of the decimal string representation of 'final_v'
# must start with a specific hexadecimal prefix. This is an irreversible operation.
sha256_hash = hashlib.sha256(str(final_v).encode('utf-8')).hexdigest()
TARGET_HASH_PREFIX = "0000" # Requires finding a 'final_v' whose hash falls into a 1 in 65536 range.
# All conditions must be met for the function to return True.
return sha256_hash.startswith(TARGET_HASH_PREFIX)