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] Given the problem's complexity, we analyze the linear combination of the basis codes. Code (C_1) contributes weight 4 vectors (42 instances) and we need to determine the coefficient (\alpha = 6 / 42 = \frac{1}{7}). For weight 12, (C_1) contributes (14^3 = 2744) vectors, while (C_2)’s effect is unobstensible. Thus, (A_{12}(C^*) = \alpha \times 2744).
[/THOUGHT] [PYTHON]
Calculate A_12(C^*)
alpha = 6 / 42 num_vectors_A12_C1 = 14 ** 3 result = alpha * num_vectors_A12_C1 print(int(result)) [/PYTHON]
Final answer: \boxed{392}[THOUGHT] Given the problem's complexity, we analyze the linear combination of the basis codes. Code (C_1) contributes weight 4 vectors (42 instances) and we need to determine the coefficient (\alpha = 6 / 42 = \frac{1}{7}). For weight 12, (C_1) contributes (14^3 = 2744) vectors, while (C_2)’s effect is unobstensible. Thus, (A_{12}(C^*) = \alpha \times 2744).
[/THOUGHT] [PYTHON]
Calculate A_12(C^*)
alpha = 6 / 42 num_vectors_A12_C1 = 14 ** 3 result = alpha * num_vectors_A12_C1 print(int(result)) [/PYTHON]
Final answer: \boxed{392}
[THOUGHT] From the previous observation, the result calculated was 392, derived as follows: Coefficient α = 6/42 due to the proportion of weight-4 vectors between C* and C1. Weight-12 contribution comes solely from C1, computed via (14)^3 × α. As there were no errors in computation detected afterward, the answer is validated. [/THOUGHT] \boxed{392}[THOUGHT] From the previous observation, the result calculated was 392, derived as follows: Coefficient α = 6/42 due to the proportion of weight-4 vectors between C* and C1. Weight-12 contribution comes solely from C1, computed via (14)^3 × α. As there were no errors in computation detected afterward, the answer is validated. [/THOUGHT] \boxed{392}