def mystery(x):
if not isinstance(x, str) or len(x) > 20:
return False
# Restrict characters to digits, operators, parentheses, and spaces
if not set(x).issubset(set("0123456789+-*/() ")):
return False
# The digits in the string must exactly match this descending sequence
if "".join(c for c in x if c.isdigit()) != "987654321":
return False
try:
# The math expression must evaluate exactly to 100
if eval(x) == 100:
# It must use all four basic arithmetic operators at least once
return set("+-*/").issubset(set(x))
except Exception:
pass
return False
def mystery(x):
if not (isinstance(x, str) and len(x) == 10 and x.isdigit()):
return False
# Each digit at index i must correctly represent the frequency
# of the digit i within the entire string.
if any(len([c for c in x if c == str(i)]) != int(d) for i, d in enumerate(x)):
return False
# A positional summation constraint using bitwise shifts
# This further validates the specific arrangement of digits.
return sum(int(d) << (i * 4) for i, d in enumerate(x)) == 16777510
def mystery(x):
if not isinstance(x, str) or len(x) > 100:
return False
# Only a very restricted set of characters is allowed!
allowed = set(" ()%':abdlmrx")
if not set(x).issubset(allowed):
return False
try:
# The expression must evaluate to exactly itself!
# (A pure expression Quine)
return eval(x) == x
except Exception:
return False
def mystery(x):
if not isinstance(x, int) or not (0 <= x < 2**32):
return False
try:
# Interpret the number as a sequence of 4 bytes
b = x.to_bytes(4, 'big')
# Constraint 1: A quadratic relation between the first two bytes
cond1 = (b[0]**2 + b[1]**2 == 1874)
# Constraint 2: Bitwise manipulation of the remaining bytes
cond2 = ((b[2] << 2) ^ (b[3] >> 1) == 165)
# Constraint 3: XOR differences between specific byte pairs
cond3 = ((b[0] ^ b[2]) + (b[1] ^ b[3]) == 18)
# Constraint 4: A crossover summation-XOR check
cond4 = ((b[0] + b[3]) ^ (b[1] + b[2]) == 24)
return cond1 and cond2 and cond3 and cond4
except Exception:
return False
def mystery(x):
if not isinstance(x, str) or len(x) > 65:
return False
# Strictly limit the allowed characters
allowed = set(" lambda:()tfx")
if not set(x).issubset(allowed):
return False
try:
# Evaluate the pure lambda expression
func = eval(x)
# func must be a valid Church numeral!
# It should apply the increment function exactly 65,536 times.
return func(lambda i: i + 1)(0) == 65536
except Exception:
return False
def mystery(x):
if not (isinstance(x, str) and len(x) == 12 and x.isdigit()):
return False
try:
# Split the 12-digit string into three 4-digit integers
a, b, c = (int(x[i:i+4]) for i in range(0, 12, 4))
# Constraint 1: A linear sum relationship
cond1 = (a + b + c == 14580)
# Constraint 2: A bitwise XOR checksum
cond2 = (a ^ b ^ c == 13954)
# Constraint 3: A cyclic product-sum modulo check
cond3 = ((a*b + b*c + c*a) % 10000 == 8133)
# Constraint 4: Digital sum of the entire string
cond4 = (sum(map(int, x)) == 36)
# Constraint 5: Positional ordering for a unique solution
cond5 = (a < b < c)
# Constraint 6: Complexity check on the character set
cond6 = (len(set(x)) == 7)
return cond1 and cond2 and cond3 and cond4 and cond5 and cond6
except Exception:
return False
def mystery(x):
if not isinstance(x, str) or len(x) > 30:
return False
# Strictly restricted character set
allowed = set("[]+*0123456789 ")
if not set(x).issubset(allowed):
return False
try:
res = eval(x)
if type(res) is not list:
return False
# Recursive signature function
def sig(l):
if type(l) is not list: return 0
return 1 + sum(sig(i) * (k+1) for k, i in enumerate(l))
return sig(res) == 8047404
except Exception:
return False
def mystery(x):
if not (isinstance(x, str) and len(x) == 10 and x.isdigit()):
return False
# The string must be a permutation of all digits from 0 to 9
if sorted(x) != list("0123456789"):
return False
p = [int(d) for d in x]
# Property 1: The permutation must form a single cycle of length 10
# (a Hamiltonian cycle in the graph of the permutation).
curr, mask = 0, 0
for _ in range(10):
curr = p[curr]
mask |= (1 << curr)
if mask != 1023: # 1023 is binary 1111111111
return False
# Property 2: A system of arithmetic and bitwise constraints
# that further restrict the possible permutations.
cond1 = sum(i * p[i] for i in range(10)) == 130
cond2 = sum((i + 1) ^ (p[i] + 1) for i in range(10)) == 78
cond3 = sum(p[i] * p[(i + 1) % 10] for i in range(10)) == 259
return cond1 and cond2 and cond3
def mystery(x):
if not isinstance(x, str) or len(x) > 30:
return False
# Stack machine with a very limited instruction set
allowed = set("123+*dp")
if not set(x).issubset(allowed):
return False
stack = []
try:
for op in x:
if op in "123":
stack.append(int(op))
elif op == '+':
stack.append(stack.pop() + stack.pop())
elif op == '*':
stack.append(stack.pop() * stack.pop())
elif op == 'd':
stack.append(stack[-1])
elif op == 'p':
b = stack.pop()
a = stack.pop()
# Prevent Memory/Time DOS from astronomically large exponents
if a > 100000 or b > 100000:
return False
stack.append(a ** b)
# The result must be exactly this enormous number
target = 2**65536 + 3**65536 + 1
return len(stack) == 1 and stack[0] == target
except Exception:
return False
def mystery(x):
if not (isinstance(x, str) and len(x) == 12 and x.isdigit()):
return False
try:
# Segment the string into three 4-digit blocks
segments = [x[i:i+4] for i in range(0, 12, 4)]
a, b, c = map(int, segments)
except ValueError:
return False
# Constraint 1: The segments must satisfy both linear and bitwise sums.
# The property (a + b + c) == (a ^ b ^ c) implies that the binary
# representations of a, b, and c have no overlapping set bits.
if (a + b + c) != 12345 or (a ^ b ^ c) != 12345:
return False
# Constraint 2: Each segment must have a Hamming weight of exactly 2.
if not all(bin(n).count('1') == 2 for n in [a, b, c]):
return False
# Constraint 3: Segments must be in strictly increasing order.
if not (a < b < c):
return False
# Constraint 4: The total sum of all individual digits must be 42.
if sum(map(int, x)) != 42:
return False
# Constraint 5: A specific digit frequency check to ensure uniqueness.
return x.count('4') == 3