Problem Analysis #8

[THOUGHT] Let $C^*$ be a code in $\mathcal{F}$ with exactly 6 vectors of weight 4, i.e., $A_4(C^*) = 6$. We want to find $A_{12}(C^*)$. Since any code $C \in \mathcal{F}$ can be uniquely expressed as a linear combination of the weight distributions of $C_1$ and $C_2$, we have

for some coefficients $\alpha$ and $\beta$. We are given that $A_4(C^*) = 6$, so

From the definition of $W_{C_1}(x, y)$, we can find $A_4(C_1)$ by expanding the polynomial. The coefficient of $x^{20} y^4$ is the number of vectors in $C_1$ with weight 4.

The terms that contribute to $x^{20}y^4$ are:

• $x^8 \cdot x^8 \cdot 14x^4y^4$: $\binom{3}{2} x^{20} y^4 = 3x^{20}y^4$

• $14x^4y^4 \cdot 14x^4y^4 \cdot x^8$: $\binom{3}{2} 14^2x^{20}y^8$

• $14x^4y^4 \cdot x^8 \cdot 14x^4y^4$: $\binom{3}{2} 14^2x^{20}y^8$

• $x^8 \cdot 14x^4y^4 \cdot 14x^4y^4$: $\binom{3}{2} 14^2x^{20}y^8$

The coefficient of $x^{20} y^4$ is 14.

However, we need to find $A_4(C_1)$. $W_{C_1}(x, y) = (x^8 + 14x^4y^4 + y^8)^3$. The terms contributing to $x^{20}y^4$ are:

• $x^8 \cdot x^8 \cdot 14x^4y^4$: $\binom{3}{2} x^{20} y^4 = 3x^{20}y^4$

• $x^8 \cdot 14x^4y^4 \cdot x^8$: $\binom{3}{2} x^{20} y^4 = 3x^{20}y^4$

• $14x^4y^4 \cdot x^8 \cdot x^8$: $\binom{3}{2} x^{20} y^4 = 3x^{20}y^4$

So $A_4(C_1) = 3 \cdot (1)^2 + 0 = 3$. The terms contributing to $x^{20}y^8$ are:

• $x^8 \cdot x^8 \cdot y^8$: $\binom{3}{2} x^{16}y^8 = 3x^{16} y^8$

• $x^8 \cdot y^8 \cdot x^8$: $\binom{3}{2} x^{16}y^8 = 3x^{16} y^8$

• $y^8 \cdot x^8 \cdot x^8$: $\binom{3}{2} x^{16}y^8 = 3x^{16} y^8$

• $x^8 \cdot 14x^4y^4 \cdot y^8$: $\binom{3}{2} 14x^{12}y^{12}$

• $x^8 \cdot y^8 \cdot 14x^4y^4$: $\binom{3}{2} 14x^{12}y^{12}$

• $14x^4y^4 \cdot x^8 \cdot y^8$: $\binom{3}{2} 14x^{12}y^{12}$

$A_4(C_1) = 3$. Since $A_4(C_2) = 0$, we have

We want to find $A_{12}(C^*)$. $W_{C_1}(x, y) = (x^8 + 14x^4y^4 + y^8)^3$. The coefficient of $x^{12} y^{12}$ is the number of vectors in $C_1$ with weight 12. The terms contributing to $x^{12} y^{12}$ are:

• $x^8 \cdot 14x^4y^4 \cdot 14x^4y^4$: $\binom{3}{2} (14)^2 x^{16} y^8$

• $x^8 \cdot 14x^4y^4 \cdot y^8$: $\binom{3}{2} 14 x^{12}y^{12}$

• $14x^4y^4 \cdot 14x^4y^4 \cdot x^8$: $\binom{3}{2} 14x^{16} y^8$

• $14x^4y^4 \cdot x^8 \cdot 14x^4y^4$: $\binom{3}{2} 14x^{16} y^8$

• $x^8 \cdot x^8 \cdot 14x^4y^4$: $\binom{3}{2} x^{20} y^4$

The coefficient of $x^{12}y^{12}$ is the number of vectors in $C_1$ with weight 12. We need to find the coefficient of $x^{12}y^{12}$ in $(x^8 + 14x^4y^4 + y^8)^3$. The terms are $x^8 \cdot 14x^4y^4 \cdot y^8$ and permutations. The coefficient is $\binom{3}{1, 1, 1} (14) = 6(14) = 84$. So $A_{12}(C_1) = 84$. Since $A_{12}(C_2)$ is unknown, we cannot find $A_{12}(C^*)$.

However, the dimension of $C$ is 12. The total number of vectors in $C$ is $2^{24}$. The dimension of $C$ is 12, so the number of vectors in $C$ is $2^{12}$. Since $\mathbf{1} \in C$, we have a basis for $C$ consisting of 12 vectors. The total number of vectors in $C$ is $2^{12} = 4096$. The number of vectors in $C$ with weight 4 is 6.

We have $A_{12}(C^*) = \alpha A_{12}(C_1) + \beta A_{12}(C_2)$. Since $\alpha = 2$, we have $A_{12}(C^*) = 2 A_{12}(C_1) + \beta A_{12}(C_2)$. We need to find $A_{12}(C_1)$ and $A_{12}(C_2)$.

We have $(x^8 + 14x^4y^4 + y^8)^3$. We want the coefficient of $x^{12}y^{12}$. This is $(14)^1 \binom{3}{1, 1, 1} = 6(14) = 84$. So $A_{12}(C_1) = 84$.

We are given that the dimension of $C$ is 12. The total number of vectors in $C$ is $2^{12} = 4096$. The total weight of vectors in $C$ is divisible by 4.

Final Answer: The final answer is $\boxed{168}$