Let . Consider the family of all linear subspaces such that
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;
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every codeword has even Hamming weight;
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is self-orthogonal:
What is ?
[THOUGHT] The family (\mathcal{F}) consists of all 4-dimensional self-dual linear codes in (\mathbb{F}_2^8), which automatically satisfy the even weight condition. To count them, count the number of ordered orthogonal bases (4 linearly independent even-weight vectors that are pairwise orthogonal, including self-orthogonal) and divide by (|\mathrm{GL}(4,2)| = 20160). Represent vectors as integers 0-255. Precompute parities and dot products. Use recursion to build bases, checking orthogonality and linear independence via span generation with XOR. [/THOUGHT] [PYTHON] par = [0] * 256 for i in range(256): par[i] = bin(i).count('1') % 2
even_nonzero = [i for i in range(256) if i > 0 and par[i] == 0]
dots = [[par[i & j] for j in range(256)] for i in range(256)]
def generate_span(basis): span = {0} for vec in basis: span = {s ^ t for s in span for t in (0, vec)} return span
total_ordered = [0]
def recurse(current_basis, depth): if depth == 4: total_ordered[0] += 1 return prev_span = generate_span(current_basis) for v in even_nonzero: if all(dots[v][b] == 0 for b in current_basis) and v not in prev_span: recurse(current_basis + [v], depth + 1)
recurse([], 0)
gl4 = 15 * 14 * 12 * 8 num_f = total_ordered[0] // gl4 print(num_f) [/PYTHON]
[THOUGHT] The Python code counted the number of ordered bases and divided by |GL(4,2)| = 20160, resulting in 135, which is the size of (\mathcal{F}). [/THOUGHT] \boxed{135}