Let be the binary linear code
How many binary linear codes are there which are equivalent (under coordinate permutations) to ?
[THOUGHT] To count codes equivalent to (C) under coordinate permutations, compute the orbit size of (C) under (S_{16}). By orbit–stabilizer, this is (16!/|\mathrm{Aut}(C)|). Here (\mathrm{Aut}(C)) consists of permutations that send each of the 8 coordinate-pairs to a (possibly different) coordinate-pair, with an optional swap inside each pair, so (|\mathrm{Aut}(C)|=2^8\cdot 8!). I will compute (16!/(2^8\cdot 8!)) exactly with Python. [/THOUGHT] [PYTHON] import math from fractions import Fraction
val = math.factorial(16) // (28 * math.factorial(8)) val2 = Fraction(math.factorial(16), 28 * math.factorial(8)) print(val) print(val2) print(val == val2) [/PYTHON][THOUGHT] To count codes equivalent to (C) under coordinate permutations, compute the orbit size of (C) under (S_{16}). By orbit–stabilizer, this is (16!/|\mathrm{Aut}(C)|). Here (\mathrm{Aut}(C)) consists of permutations that send each of the 8 coordinate-pairs to a (possibly different) coordinate-pair, with an optional swap inside each pair, so (|\mathrm{Aut}(C)|=2^8\cdot 8!). I will compute (16!/(2^8\cdot 8!)) exactly with Python. [/THOUGHT] [PYTHON] import math from fractions import Fraction
val = math.factorial(16) // (28 * math.factorial(8)) val2 = Fraction(math.factorial(16), 28 * math.factorial(8)) print(val) print(val2) print(val == val2) [/PYTHON]
[THOUGHT] Using orbit–stabilizer, the number of distinct codes equivalent to (C) under coordinate permutations is (16!/(2^8\cdot 8!)), which Python evaluated as 2027025. [ \boxed{2027025} ]