Problem Analysis #8

[THOUGHT] The problem asks for the number of linear subspaces $C \subseteq \{0,1\}^8$ of dimension 4 that are self-orthogonal. Self-orthogonal means $\mathbf{u} \cdot \mathbf{v} = 0$ for all $\mathbf{u}, \mathbf{v} \in C$. This implies $\mathbf{u} \cdot \mathbf{u} = 0$, so every codeword has even weight. Thus, we need to count the number of self-dual codes of length 8. I will use Python to iterate through all 4-dimensional subspaces of $\{0,1\}^8$ and count those that are self-orthogonal. I will represent subspaces using their Reduced Row Echelon Form (RREF) generator matrices to ensure each subspace is counted exactly once. The number of such subspaces is the Gaussian binomial coefficient $\binom{8}{4}_2 = 200,787$. This is small enough to iterate.

Algorithm:

1. Iterate over all pivot column combinations $c_1 < c_2 < c_3 < c_4$ from $\{0, \dots, 7\}$.
2. For each pivot set, identify the free parameter positions $(i, j)$ where column $j$ is not a pivot and $j > c_i$.
3. Generate all binary matrices corresponding to these free parameters.
4. For each matrix, check if the rows are pairwise orthogonal (including self-orthogonality).
   • Since the dot product is bilinear, checking the basis vectors (rows) is sufficient.
   • Check $row_i \cdot row_k \equiv 0 \pmod 2$ for $0 \le i \le k \le 3$.
5. Sum the counts.

I will also calculate the theoretical value using the formula $\prod_{i=1}^{n/2-1} (2^i + 1)$ for comparison. [/THOUGHT] [PYTHON] import itertools

def solve(): n = 8 k = 4

# Theoretical value
# Formula for number of self-dual codes of length n
# Prod_{i=1}^{n/2 - 1} (2^i + 1)
theoretical = 1
for i in range(1, n // 2):
    theoretical *= (2**i + 1)
print(f"Theoretical prediction: {theoretical}")

count = 0

# Iterate over all choices of pivot columns
for pivots in itertools.combinations(range(n), k):
    # pivots is a tuple like (0, 1, 2, 3)
    
    # Identify non-pivot columns
    pivot_set = set(pivots)
    non_pivots = [c for c in range(n) if c not in pivot_set]
    
    # Identify free positions: (row_idx, col_idx)
    # Row i has pivot at pivots[i].
    # Free entries are in non-pivot columns j such that j > pivots[i].
    free_positions = []
    for r in range(k):
        pivot_col = pivots[r]
        for c in non_pivots:
            if c > pivot_col:
                free_positions.append((r, c))
    
    num_free = len(free_positions)
    
    # Iterate over all values for free positions
    # Using itertools.product for bit combinations
    for bits in itertools.product([0, 1], repeat=num_free):
        # Construct the matrix
        # Initialize with 0s
        matrix = [[0] * n for _ in range(k)]
        
        # Set pivots
        for r in range(k):
            matrix[r][pivots[r]] = 1
        
        # Set free variables
        for idx, (r, c) in enumerate(free_positions):
            matrix[r][c] = bits[idx]
        
        # Check self-orthogonality
        # We need row_i dot row_j == 0 for all i <= j
        valid = True
        for i in range(k):
            for j in range(i, k):
                dot_prod = 0
                for x in range(n):
                    dot_prod += matrix[i][x] * matrix[j][x]
                if dot_prod % 2 != 0:
                    valid = False
                    break
            if not valid:
                break
        
        if valid:
            count += 1
            
print(f"Count found by enumeration: {count}")

solve() [/PYTHON]

The problem asks for the number of linear subspaces $C \subseteq \{0,1\}^8$ satisfying three conditions:

1. $\dim(C) = 4$.
2. Every codeword in $C$ has even Hamming weight.
3. $C$ is self-orthogonal, i.e., $\mathbf{u} \cdot \mathbf{v} = 0$ for all $\mathbf{u}, \mathbf{v} \in C$.

First, we analyze the conditions. In the field $\mathbb{F}_2$, the dot product of a vector with itself is $\mathbf{u} \cdot \mathbf{u} = \sum u_i^2 = \sum u_i \pmod 2$, which is equal to the Hamming weight of $\mathbf{u}$ modulo 2. The condition that $C$ is self-orthogonal implies that for all $\mathbf{u} \in C$, $\mathbf{u} \cdot \mathbf{u} = 0$. This means the Hamming weight of every codeword is even. Thus, condition (2) is a direct consequence of condition (3) and is redundant.

Since $\dim(C) = 4$ and the ambient space has dimension 8, a self-orthogonal subspace $C$ satisfies $C \subseteq C^\perp$. Since $\dim(C^\perp) = 8 - \dim(C) = 4$, we must have $C = C^\perp$. Such codes are called self-dual codes.

The number of self-dual binary codes of length $n$ (where $n$ is even) is given by the formula: [ N(n) = \prod_{i=1}^{n/2 - 1} (2^i + 1) ] For $n=8$, the number is: [ N(8) = \prod_{i=1}^{3} (2^i + 1) = (2^1 + 1)(2^2 + 1)(2^3 + 1) = 3 \times 5 \times 9 = 135 ]

The Python code verified this by enumerating all 4-dimensional subspaces in reduced row echelon form and counting those that are self-orthogonal. The code output was 135.

\boxed{135}