import base64
def mystery(x):
code = base64.b64decode('Nio3').decode('utf-8')
return eval(code) == x
def mystery(x):
# Input must be a 4-letter alphabetic string
if not isinstance(x, str) or len(x) != 4:
return False
if not x.isalpha():
return False
# Convert characters to ASCII values
vals = [ord(c) for c in x]
a, b, c, d = vals
# A system of equations combining bitwise XOR and arithmetic addition.
# This combination disrupts standard linear algebra techniques (like Gaussian elimination),
# requiring a constraint solver, brute force, or careful bitwise analysis.
# Constraint 1: Relationship between 1st and 2nd char
cond1 = (a ^ b) == 44
# Constraint 2: Relationship between 2nd and 3rd char
cond2 = (b ^ c) == 21
# Constraint 3: Relationship between 3rd and 4th char
cond3 = (c ^ d) == 28
# Constraint 4: The arithmetic sum of all characters
cond4 = (a + b + c + d) == 394
return cond1 and cond2 and cond3 and cond4
def mystery(x):
if not isinstance(x, int) or x <= 0:
return False
return (x % 101 == 42) and (x % 103 == 99) and (x % 107 == 5) and (x % 109 == 7)
def mystery(x):
# Input must be a 4-character string
if not isinstance(x, str) or len(x) != 4:
return False
# Convert characters to their ASCII integer values
a, b, c, d = [ord(char) for char in x]
# A system of linear constraints
# These look like random equations, but they form a matrix multiplication
# M * [a, b, c, d]^T = Target
eq1 = (2*a + b + 3*c + d) == 703
eq2 = (4*a + 3*b + 7*c + 4*d) == 1861
eq3 = (2*a + 4*b + 9*c + 8*d) == 2518
eq4 = (8*a + 5*b + 19*c + 13*d) == 4767
return eq1 and eq2 and eq3 and eq4
def mystery(x):
if not isinstance(x, str) or len(x) != 4:
return False
vals = [ord(c) for c in x]
a, b, c, d = vals
cond1 = (a * b) == 10088
cond2 = (b + c) == 211
cond3 = (c * d) == 11400
cond4 = (a + d) == 204
return cond1 and cond2 and cond3 and cond4
def mystery(x):
# Input must be a 5-letter string
if not isinstance(x, str) or len(x) != 5:
return False
# Convert characters to ASCII values
a, b, c, d, e = [ord(char) for char in x]
# A system of non-linear equations combining arithmetic and bitwise operations.
# This creates a dependency chain that is difficult to invert using standard linear algebra,
# but solvable via constraint programming (e.g. Z3) or careful algebraic deduction.
# Constraint 1: Non-linear interaction of the first four characters
# a*b - c*d
check1 = (a * b - c * d) == -4816
# Constraint 2: Non-linear interaction of the last four characters
# b*c - d*e
check2 = (b * c - d * e) == 1636
# Constraint 3: Constraint on the sum of characters at odd indices (0, 2, 4)
# (a + c + e)^2
check3 = (a + c + e)**2 == 77284
# Constraint 4: Bitwise mixing of the outer characters and the inner neighbor pair
# (a XOR e) * (b XOR d)
check4 = (a ^ e) * (b ^ d) == 1008
# Constraint 5: Simple arithmetic sum of all characters
check5 = (a + b + c + d + e) == 498
return check1 and check2 and check3 and check4 and check5
def mystery(x):
if not isinstance(x, str) or len(x) != 6:
return False
if not x.isalpha():
return False
a, b, c, d, e, f = [ord(char) for char in x]
check1 = (a * b - c * d) == -1780
check2 = (b * c - d * e) == 1098
check3 = (c * d - e * f) == 3976
check4 = (a ^ d) * (b ^ e) * (c ^ f) == 7750
check5 = (a + b + c) ** 2 == 123201
check6 = (d + e + f) ** 2 == 109561
return check1 and check2 and check3 and check4 and check5 and check6
def mystery(x):
if not isinstance(x, str) or len(x) != 6:
return False
# Extract ASCII values
v = [ord(c) for c in x]
# Constraint 1: Linear - Sum of values
if sum(v) != 642:
return False
# Constraint 2: Linear - Weighted Sum (1-based index)
if sum((i + 1) * val for i, val in enumerate(v)) != 2301:
return False
# Constraint 3: Linear - Alternating Sum (+ - + - + -)
if sum(val * ((-1)**i) for i, val in enumerate(v)) != -28:
return False
# Constraint 4: Quadratic - Sum of Squares
if sum(val**2 for val in v) != 69734:
return False
# Constraint 5: Quadratic - Symmetric Product
# Products of characters at symmetric positions (0&5, 1&4, 2&3)
if v[0]*v[5] + v[1]*v[4] + v[2]*v[3] != 34295:
return False
# Constraint 6: Non-linear - Sum of Pairwise XOR
# XOR sums of adjacent disjoint pairs
if (v[0] ^ v[1]) + (v[2] ^ v[3]) + (v[4] ^ v[5]) != 70:
return False
return True
def mystery(x):
if not isinstance(x, str) or len(x) != 7:
return False
if not x.isalpha():
return False
a, b, c, d, e, f, g = [ord(char) for char in x]
check1 = (a * b - c * d) == -1780
check2 = (b * c - d * e) == 1098
check3 = (c * d - e * f) == 3976
check4 = (d * e - f * g) == 1662
check5 = (a ^ d) * (b ^ e) * (c ^ f) * (d ^ g) == 62000
check6 = (a + b + c + d + e + f + g) == 796
check7 = (a**2 + b**2 + c**2 + d**2 + e**2 + f**2 + g**2) == 90862
return check1 and check2 and check3 and check4 and check5 and check6 and check7
def mystery(x):
# Input must be an 8-letter alphabetic string
if not isinstance(x, str) or len(x) != 8:
return False
if not x.isalpha():
return False
v = [ord(c) for c in x]
# A system of 8 constraints for 8 variables (characters).
# The system mixes linear algebra, quadratic equations, and bitwise operations.
# This structure (Mixed Boolean-Arithmetic) resists standard Gaussian elimination
# and simple algebraic inversion.
# Constraint 1: Simple Sum
if sum(v) != 841:
return False
# Constraint 2: Weighted Sum (0-based index weighting)
if sum(i * val for i, val in enumerate(v)) != 3006:
return False
# Constraint 3: Sum of Squares (Quadratic)
if sum(val**2 for val in v) != 89577:
return False
# Constraint 4: Alternating Product Sum (Quadratic/Bilinear)
# v0*v1 - v2*v3 + v4*v5 - v6*v7
term1 = v[0] * v[1]
term2 = v[2] * v[3]
term3 = v[4] * v[5]
term4 = v[6] * v[7]
if term1 - term2 + term3 - term4 != -1433:
return False
# Constraint 5: Product of XOR Sums (Non-linear bitwise mixing)
# Separates indices into evens (0,2,4,6) and odds (1,3,5,7)
xor_even = v[0] ^ v[2] ^ v[4] ^ v[6]
xor_odd = v[1] ^ v[3] ^ v[5] ^ v[7]
if xor_even * xor_odd != 86:
return False
# Constraint 6: Linear Symmetry of outer pairs
# (v0 + v7) - (v1 + v6)
if (v[0] + v[7]) - (v[1] + v[6]) != -26:
return False
# Constraint 7: Linear Symmetry of inner pairs
# (v2 + v5) - (v3 + v4)
if (v[2] + v[5]) - (v[3] + v[4]) != 13:
return False
# Constraint 8: Sum of pairwise AND (Bitwise non-linearity)
# (v0 & v1) + (v2 & v3) + (v4 & v5) + (v6 & v7)
and_sum = (v[0] & v[1]) + (v[2] & v[3]) + (v[4] & v[5]) + (v[6] & v[7])
if and_sum != 377:
return False
return True