def mystery(x):
if not isinstance(x, str) or not x.isdigit():
return False
n = int(x)
if n <= 0:
return False
# Condition 1: The number of unique digits in the string x must be a prime number.
unique_digits = set(x)
num_unique_digits = len(unique_digits)
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
if not is_prime(num_unique_digits):
return False
# Condition 2: The length of the string x must be a power of 2.
s_length = len(x)
if not (s_length > 0 and (s_length & (s_length - 1) == 0)):
return False
# Condition 3: The integer n, when converted to its binary representation, must be a palindrome.
binary_n = bin(n)[2:] # Get binary string without '0b' prefix
if not (binary_n == binary_n[::-1]):
return False
return True
def mystery(x):
import unicodedata
# Accept only text input; numbers are rejected.
if not isinstance(x, str):
return False
# Normalize to reduce "look-alike" tricks.
x = unicodedata.normalize("NFKC", x)
# Basic shape constraints (printable ASCII, exact length).
if len(x) != 16:
return False
for ch in x:
o = ord(ch)
if o < 0x20 or o > 0x7E:
return False
# --- Obfuscated target reconstruction ---
# 16 encrypted bytes stored as 4x32-bit words.
parts = (
0xFFE47DD3,
0xDB4725EE,
0xE47B95B6,
0x5E4BF5F4,
)
blob = b"".join(p.to_bytes(4, "big") for p in parts)
# Deterministic per-position mask (keystream) + fixed whitening constant.
def _mask(i):
return ((i * 73 + 41) & 0xFF) ^ 0xA5
secret_bytes = bytes((b ^ _mask(i)) for i, b in enumerate(blob))
# Ensure the hidden secret is plain ASCII.
try:
secret_bytes.decode("ascii")
except Exception:
return False
# --- Comparison (intentionally not written as `x == secret`) ---
xb = x.encode("ascii", "strict")
if len(xb) != len(secret_bytes):
return False
diff = 0
for a, b in zip(xb, secret_bytes):
diff |= (a ^ b)
if diff != 0:
return False
# A couple of extra constraints that are redundant if the above matches,
# but add noise for anyone trying to shortcut by inspection.
s1 = 0
s2 = 0
for i, v in enumerate(xb):
s1 = (s1 + ((v + i) * (i + 17))) & 0xFFFFFFFF
s2 ^= ((v << (i % 5)) | (v >> (8 - (i % 5)))) & 0xFF
t1 = 0
t2 = 0
for i, v in enumerate(secret_bytes):
t1 = (t1 + ((v + i) * (i + 17))) & 0xFFFFFFFF
t2 ^= ((v << (i % 5)) | (v >> (8 - (i % 5)))) & 0xFF
return (s1 == t1) and (s2 == t2)
def mystery(x):
# Condition 0: Basic type and value checks
if not isinstance(x, str) or not x.isdigit():
return False
n = int(x)
if n <= 0:
return False
# Condition 1: The length of the string x must be a prime number.
s_length = len(x)
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
if not is_prime(s_length):
return False
# Condition 2: The "digital root" of the integer n must be 4.
# The digital root is the single-digit value obtained by iteratively summing
# the digits of a number until a single digit remains.
# For positive integers, this is equivalent to n % 9 (or 9 if n % 9 == 0).
if n % 9 != 4:
return False
# Condition 3: The digits of x must strictly alternate in parity
# (i.e., Odd-Even-Odd-Even... or Even-Odd-Even-Odd...).
parities = []
for char_digit in x:
digit = int(char_digit)
parities.append(digit % 2) # 0 for even, 1 for odd
if len(parities) > 1:
is_alternating = True
for i in range(len(parities) - 1):
if parities[i] == parities[i+1]:
is_alternating = False
break
if not is_alternating:
return False
# Single-digit numbers (len=1) are trivially alternating.
# Condition 4: The sum of the digits of n must be exactly one less
# than the sum of the digits of n + 1.
def sum_digits(num_val):
s = 0
temp_num = num_val
while temp_num > 0:
s += temp_num % 10
temp_num //= 10
return s
if not (sum_digits(n) == sum_digits(n + 1) - 1):
return False
return True
def mystery(x):
import unicodedata
# --- Input gatekeeping ---
if not isinstance(x, str):
return False
x = unicodedata.normalize("NFKC", x)
# Must be exactly 20 visible ASCII characters (no spaces, no control chars).
if len(x) != 20:
return False
for ch in x:
o = ord(ch)
if o < 0x21 or o > 0x7E:
return False
xb = x.encode("ascii", "strict")
# --- Hidden target reconstruction (no cleartext secret in source) ---
# 20 bytes packed as 5 big-endian 32-bit words, but also permuted.
words = (
0xD3260FB5,
0x068571B8,
0xA677D0F5,
0x229D5C9C,
0x9AC43BA0,
)
blob = b"".join(w.to_bytes(4, "big") for w in words) # 20 bytes (permuted)
# Undo a position permutation: stored[j] = C[(j*7) mod 20]
# Inverse exists since 7 is invertible mod 20 (inverse = 3), so C[i] = stored[(i*3) mod 20]
C = bytes(blob[(i * 3) % 20] for i in range(20))
def mask(i: int) -> int:
return (29 * i * i + 101 * i + 17) & 0xFF
def k(i: int) -> int:
return (0xA7 + 9 * i) & 0xFF
secret_bytes = bytes(((C[i] ^ k(i)) - mask(i)) & 0xFF for i in range(20))
# Secret must be ASCII; if not, reject.
try:
secret_bytes.decode("ascii")
except Exception:
return False
# --- Constant-time style equality check ---
diff = 0
for a, b in zip(xb, secret_bytes):
diff |= (a ^ b)
if diff != 0:
return False
# --- Extra redundant constraints (noise) ---
s_mod257 = 0
s_xor = 0
for b in xb:
s_mod257 = (s_mod257 + b) % 257
s_xor ^= b
# These constants are consistent with the hidden secret.
return (s_mod257 == 74) and (s_xor == 77)
def mystery(x):
# Condition 0: Basic type and content checks
if not isinstance(x, str):
return False
vowels = {'a', 'e', 'i', 'o', 'u'}
# Check if all characters are lowercase English letters and apply parity pattern
char_sum = 0
for i, char in enumerate(x):
if not ('a' <= char <= 'z'):
return False # Not a lowercase English letter
char_sum += ord(char)
# Condition 3: Character type pattern by index parity
# Characters at even indices must be vowels.
# Characters at odd indices must be consonants.
if i % 2 == 0: # Even index (0, 2, 4, ...)
if char not in vowels:
return False
else: # Odd index (1, 3, 5, ...)
if char in vowels:
return False
# An empty string would pass the loop but fail subsequent checks naturally.
# However, for clarity and to avoid edge cases with prime/perfect square logic,
# we explicitly reject it as it cannot meet the criteria meaningfully.
if len(x) == 0:
return False
# Condition 1: The length of the string x must be a prime number.
s_length = len(x)
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
if not is_prime(s_length):
return False
# Condition 2: The sum of the ASCII values of all characters in x
# must be a perfect square (e.g., 1, 4, 9, 16, 25, ...).
# (char_sum was calculated in the loop above)
# Minimum ASCII value for 'a' is 97, so char_sum for non-empty string is always positive.
sqrt_char_sum = int(char_sum**0.5)
if not (sqrt_char_sum * sqrt_char_sum == char_sum):
return False
return True
def mystery(x):
import unicodedata
# --- Accept either int or str; canonicalize to a string ---
if isinstance(x, bool):
return False
if isinstance(x, int):
if x < 0:
return False
x = str(x)
elif not isinstance(x, str):
return False
x = unicodedata.normalize("NFKC", x)
# --- Shape constraints ---
if len(x) != 24:
return False
for ch in x:
o = ord(ch)
# visible ASCII, no spaces/control
if o < 0x21 or o > 0x7E:
return False
xb = x.encode("ascii", "strict")
# --- Hidden target reconstruction (no cleartext target in source) ---
# 24 ciphertext bytes packed as 6 big-endian 32-bit words (and permuted).
words = (
0xE0D997CB,
0x1204F575,
0x3DA7C138,
0xA936666D,
0xF2C3D686,
0xF5F1D824,
)
blob = b"".join(w.to_bytes(4, "big") for w in words) # length 24
# Stored as p[j] = c[(7*j) mod 24]; since 7 is self-inverse mod 24,
# recover original c[i] = p[(7*i) mod 24].
c = bytes(blob[(i * 7) % 24] for i in range(24))
def k(i: int) -> int:
return (((17 * i * i + 31 * i + 0x55) & 0xFF) ^ 0xC3)
secret = bytes((c[i] ^ k(i)) for i in range(24))
# --- Constant-time-ish comparison ---
diff = 0
for a, b in zip(xb, secret):
diff |= (a ^ b)
if diff != 0:
return False
# --- Extra redundant "noise" checks (consistent with the secret) ---
s_mod257 = sum(xb) % 257
s_xor = 0
for b in xb:
s_xor ^= b
return (s_mod257 == 77) and (s_xor == 0x25)
import math
def mystery(x):
# Condition 0: Basic type and content checks
if not isinstance(x, str) or not x.isdigit():
return False
n = int(x)
if n <= 0: # Ensure positive integer
return False
s_length = len(x)
# Condition 1: The length of the string x must be a prime number.
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
if not is_prime(s_length):
return False
# Helper for Condition 2
def is_fibonacci(num):
if num < 1:
return False
a, b = 0, 1
while b < num:
a, b = b, a + b
return b == num or num == 1 # Handle num=1 as a Fibonacci number
digit_sum = 0
digit_product = 1
for i, char_digit in enumerate(x):
digit = int(char_digit)
# Condition 3 (part 1): All digits must be from the set {1, 2, 4, 8}
# This is because the product of digits must be a power of 2.
if digit not in {1, 2, 4, 8}:
return False
digit_product *= digit
# Condition 4: For each digit d_i at index i (0-indexed), d_i must be strictly greater than i.
if not (digit > i):
return False
digit_sum += digit
# Condition 2: The sum of the digits of n must be a Fibonacci number.
if not is_fibonacci(digit_sum):
return False
# Condition 3 (part 2): The product of the digits of n must be a power of 2.
# A number is a power of 2 if it's positive and has only one bit set (e.g., 1, 2, 4, 8, 16, ...).
if not (digit_product > 0 and (digit_product & (digit_product - 1) == 0)):
return False
return True
def mystery(x):
import unicodedata
# --- Accept either int or str; canonicalize to a string ---
if isinstance(x, bool):
return False
if isinstance(x, int):
if x < 0:
return False
x = str(x)
elif not isinstance(x, str):
return False
x = unicodedata.normalize("NFKC", x)
# --- Shape constraints: exactly 28 visible ASCII chars (no spaces) ---
if len(x) != 28:
return False
for ch in x:
o = ord(ch)
if o < 0x21 or o > 0x7E:
return False
try:
xb = x.encode("ascii", "strict")
except Exception:
return False
# --- Hidden target reconstruction (no cleartext secret in source) ---
# 28 stored bytes packed as 7 big-endian 32-bit words.
words = (
0x2065D985,
0x5CBB116A,
0xF853AFDE,
0x65C502C4,
0xCB2562BA,
0x4A7FC1B2,
0xC6ED8295,
)
blob = b"".join(w.to_bytes(4, "big") for w in words) # 28 bytes
# Stored permutation: blob[j] = C[(11*j) mod 28]
# Invert with inv11 mod 28 = 23 => C[i] = blob[(23*i) mod 28]
C = bytes(blob[(23 * i) % 28] for i in range(28))
def k(i: int) -> int:
return (((37 * i + 101) & 0xFF) ^ 0xD6)
def add(i: int) -> int:
return ((17 * i + 53) & 0xFF)
# Decrypt: base = (C - add) mod 256; secret = base XOR k
secret = bytes((((C[i] - add(i)) & 0xFF) ^ k(i)) for i in range(28))
# Ensure reconstructed secret is printable ASCII (a bit of extra gatekeeping).
try:
s = secret.decode("ascii")
except Exception:
return False
for ch in s:
o = ord(ch)
if o < 0x21 or o > 0x7E:
return False
# --- Constant-time-ish equality check ---
diff = 0
for a, b in zip(xb, secret):
diff |= (a ^ b)
if diff != 0:
return False
# --- Redundant "noise" checks (consistent with the secret) ---
s_mod257 = sum(xb) % 257
s_xor = 0
for b in xb:
s_xor ^= b
wsum = 0
for i, b in enumerate(xb):
wsum = (wsum + (i + 1) * b) & 0xFFFF
return (s_mod257 == 26) and (s_xor == 0x65) and (wsum == 33144)
import math
def mystery(x):
# Condition 0: Basic type and length checks
if not isinstance(x, str):
return False
if len(x) != 10:
return False
digits_str = ""
letters_str = ""
for i, char in enumerate(x):
# Condition 1: Character type alternation
# Characters at even indices (0, 2, 4, 6, 8) must be digits.
# Characters at odd indices (1, 3, 5, 7, 9) must be lowercase English letters.
if i % 2 == 0: # Even index
if not char.isdigit():
return False
digits_str += char
else: # Odd index
if not ('a' <= char <= 'z'):
return False
letters_str += char
# After the loop, digits_str will contain 5 digits and letters_str will contain 5 letters.
# Condition 2: The integer N formed by concatenating the digits must be a prime number.
# N is formed from x[0], x[2], x[4], x[6], x[8]
n_digits = int(digits_str)
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
if not is_prime(n_digits):
return False
# Condition 3: The string S formed by concatenating the letters must be a palindrome.
# S is formed from x[1], x[3], x[5], x[7], x[9]
if not (letters_str == letters_str[::-1]):
return False
# Condition 4: The sum of the 1-indexed alphabetical positions of the letters in S must be 50.
# ('a' = 1, 'b' = 2, ..., 'z' = 26)
letter_pos_sum = 0
for char in letters_str:
letter_pos_sum += (ord(char) - ord('a') + 1) # 1-indexed position
if letter_pos_sum != 50:
return False
return True
def mystery(x):
import unicodedata
# --- accept either int or str; canonicalize to str ---
if isinstance(x, bool):
return False
if isinstance(x, int):
if x < 0:
return False
x = str(x)
elif not isinstance(x, str):
return False
x = unicodedata.normalize("NFKC", x)
# --- shape constraints: exactly 32 visible ASCII chars (no spaces) ---
if len(x) != 32:
return False
for ch in x:
o = ord(ch)
if o < 0x21 or o > 0x7E:
return False
try:
xb = x.encode("ascii", "strict")
except Exception:
return False
# --- hidden target reconstruction (no cleartext target in source) ---
# 32 stored bytes packed as 8 big-endian 32-bit words.
words = (
0x4C6FFCE9,
0x4AC5552F,
0x46FAAAC4,
0xB62EAB96,
0x36277435,
0x8DF0E38A,
0x80BCDB14,
0xF4062281,
)
stored = b"".join(w.to_bytes(4, "big") for w in words) # 32 bytes
# stored[j] = c[(5*j) mod 32]; invert with inv5 mod 32 = 13 => c[i] = stored[(13*i) mod 32]
c = bytes(stored[(13 * i) % 32] for i in range(32))
def add(i: int) -> int:
return (29 * i + 73) & 0xFF
def k(i: int) -> int:
t = (17 * i * i + 91 * i + 0xD3) & 0xFF
return t ^ 0xA6
secret = bytes((((c[i] - add(i)) & 0xFF) ^ k(i)) for i in range(32))
# ensure the reconstructed secret is printable ASCII (redundant, but noisy)
try:
s = secret.decode("ascii")
except Exception:
return False
for ch in s:
o = ord(ch)
if o < 0x21 or o > 0x7E:
return False
# --- constant-time-ish equality check ---
diff = 0
for a, b in zip(xb, secret):
diff |= (a ^ b)
if diff != 0:
return False
# --- extra redundant "noise" checks (consistent with the secret) ---
# simple mixed checksums to discourage shortcut reasoning
s1 = 0
s2 = 0
for i, b in enumerate(xb):
s1 = (s1 + ((b ^ (i * 131)) * (i + 17))) & 0xFFFFFFFF
s2 ^= ((b << (i & 7)) | (b >> (8 - (i & 7)))) & 0xFF
t1 = 0
t2 = 0
for i, b in enumerate(secret):
t1 = (t1 + ((b ^ (i * 131)) * (i + 17))) & 0xFFFFFFFF
t2 ^= ((b << (i & 7)) | (b >> (8 - (i & 7)))) & 0xFF
return (s1 == t1) and (s2 == t2)