Problem Analysis #8

Let $n=24$. We consider the family $\mathcal{F}$ of all linear subspaces $C \subseteq \{0,1\}^{24}$ that satisfy the following three conditions:

• The dimension of $C$ is 12.

• The Hamming weight of every vector in $C$ is a multiple of 4.

• The vector $\mathbf{1} = (1, 1, \dots, 1)$ is contained in $C$.

For any $C \in \mathcal{F}$, let $A_k(C)$ denote the number of vectors in $C$ having Hamming weight $k$. It is a known theoretical result that the weight distribution sequence $(A_0(C), A_1(C), \dots, A_{24}(C))$ for any $C \in \mathcal{F}$ can be uniquely expressed as a linear combination of the weight distributions of two fixed "basis" codes, $C_1$ and $C_2$, which are also in $\mathcal{F}$.

The basis codes are defined as follows:

• \textbf{Code $C_1$}: The weight enumerator polynomial $W_{C_1}(x, y) = \sum_{v \in C_1} x^{24-\text{wt}(v)} y^{\text{wt}(v)}$ is given by:

  $$W_{C_1}(x, y) = \left( x^8 + 14x^4y^4 + y^8 \right)^3$$

• \textbf{Code $C_2$}: This code has no vectors of weight 4 (i.e., $A_4(C_2) = 0$) and has exactly 759 vectors of weight 8 (i.e., $A_8(C_2) = 759$).

Let $C^*$ be a specific code in $\mathcal{F}$ that contains exactly 6 vectors of weight 4.

Calculate the number of vectors in $C^*$ that have weight 12.