Let . We consider the family of all linear subspaces that satisfy the following three conditions:
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The dimension of is 12.
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The Hamming weight of every vector in is a multiple of 4.
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The vector is contained in .
For any , let denote the number of vectors in having Hamming weight . It is a known theoretical result that the weight distribution sequence for any can be uniquely expressed as a linear combination of the weight distributions of two fixed "basis" codes, and , which are also in .
The basis codes are defined as follows:
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\textbf{Code }: The weight enumerator polynomial is given by:
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\textbf{Code }: This code has no vectors of weight 4 (i.e., ) and has exactly 759 vectors of weight 8 (i.e., ).
Let be a specific code in that contains exactly 6 vectors of weight 4.
Calculate the number of vectors in that have weight 12.
[THOUGHT] The weight distribution of any C in F is A_k(C) = α A_k(C1) + β A_k(C2) with α + β = 1. For C*, A_4(C*) = 6 = α * 42 + β * 0, so α = 6/42 = 1/7, β = 6/7. C2 is the extended binary Golay code with A_8(C2) = 759 confirming the identification, and its full distribution is A_0=1, A_4=0, A_8=759, A_12=2576, A_16=759, A_20=0, A_24=1. Thus A_12(C*) = (1/7)*2828 + (6/7)*2576. I will verify the expansion of C1's generating function and compute the value in Python. [/THOUGHT] [PYTHON]
Compute coefficients of (s**2 + 14*s + 1)**3
from sympy import expand, symbols s = symbols('s') p = (s**2 + 14*s + 1)**3 expanded = expand(p) coeffs = expanded.as_poly().all_coeffs()[::-1] # coeffs[0] for s^0, etc.
print("C1 coeffs for s^0 to s^6 (A0 to A24):") for i, c in enumerate(coeffs): k = 4 * i print(f"A_{k} = {c}")
Confirm C2 Golay distribution sums to 4096
A2 = [1, 0, 759, 2576, 759, 0, 1] print("\nC2 sum:", sum(A2))
Compute for C*
alpha = 1/7 beta = 6/7 A12_C1 = coeffs[3] # s^3 A12_C2 = 2576 A12_star = alpha * A12_C1 + beta * A12_C2 print("\nA12(C*):", A12_star)
Verify integer
print("As integer:", int(A12_star)) [/PYTHON]
[THOUGHT] The Python output confirms A_4(C1)=42, A_12(C1)=2828, A_12(C2)=2576. With α=1/7, β=6/7, A_12(C*)=2612 exactly. [/THOUGHT] \boxed{2612}
Explanation
The reasoning follows from the problem's structure: the weight distribution of any code (C \in \mathcal{F}) is uniquely expressed as (A_k(C) = \alpha A_k(C_1) + \beta A_k(C_2)) where (\alpha + \beta = 1). For (C^), (A_4(C^) = 6), and given (A_4(C_1) = 42) and (A_4(C_2) = 0), solve (6 = \alpha \cdot 42 + \beta \cdot 0), yielding (\alpha = 6/42 = 1/7) and (\beta = 6/7).
To find (A_{12}(C^)), use the values (A_{12}(C_1) = 2828) and (A_{12}(C_2) = 2576), both verified from the Python output. Compute (A_{12}(C^) = (1/7) \cdot 2828 + (6/7) \cdot 2576).
The Python code expanded (W_{C_1}(x, y)) by substituting (s = y^4 / x^4) into ((x^8 + 14x^4 y^4 + y^8)^3), confirming the coefficients: (A_0=1), (A_4=42), (A_8=591), (A_12=2828), etc., summing to (2^{12} = 4096). It also verified (C_2)'s distribution sums to 4096.
The computation yields ((1/7) \cdot 2828 + (6/7) \cdot 2576 = 2612), an integer matching the output, satisfying the conditions for (C^* \in \mathcal{F}).