<think> Okay, let's try to figure out how to solve this problem step by step. So, we have a specific code C* in the family F, and we need to find the number of vectors in C* that have weight 12.

First, I remember from the problem statement that any code C in F can be expressed as a linear combination of two basis codes, C1 and C2. So, the weight distribution A_k(C) for any C is a combination of A_k(C1) and A_k(C2).

Given that C* has exactly 6 vectors of weight 4, I need to find how this relates to the coefficients in the linear combination. Let's denote the coefficients as α and β such that:

A_k(C*) = α * A_k(C1) + β * A_k(C2)

We know that for k=4, A_4(C*) is 6. So,

6 = α * A_4(C1) + β * A_4(C2)

But from the problem statement, C2 has no vectors of weight 4 (A_4(C2)=0), so this simplifies to:

6 = α * A_4(C1)

So, if I can find A_4(C1), I can solve for α.

Looking at the weight enumerator polynomial for C1: W_{C1}(x,y) = (x^8 + 14x^4y^4 + y^8)^3. To find A_4(C1), which is the number of vectors in C1 with weight 4, I need to look at the coefficient of y^4 in W_{C1}(x,y).

Wait, actually, the polynomial is given as a sum over all vectors v in C1 of x^{24 - wt(v)} y^{wt(v)}. So, each term corresponds to a vector's weight. The coefficient of y^k x^{24-k} gives A_k(C1).

So, expanding W_{C1}(x,y), the coefficient of y^4 would come from terms where the total weight is 4. Let me think about how this expansion works.

The polynomial is (x^8 + 14x^4y^4 + y^8)^3. When we expand this, each term in the product will contribute to different powers of x and y.

To get a term with y^4, we need to consider all possible ways to pick terms from each of the three factors such that their exponents add up to 4 for y.

Looking at each factor: x^8 contributes y^0, 14x^4y^4 contributes y^4, and y^8 contributes y^8. So, when multiplying three factors together, the possible combinations are:

• Pick x^8 from all three factors: total y exponent is 0.
• Pick x^8 from two factors and 14x^4y^4 from one: total y exponent is 4.
• Other combinations would result in higher exponents.

So, the only way to get a term with y^4 is by picking 14x^4y^4 once and x^8 twice. The number of ways to do this is C(3,1) = 3. Each such term contributes (14)^1 * (1)^2 = 14. So, the coefficient for y^4 is 3*14 = 42.

Therefore, A_4(C1) = 42.

Going back to our equation:

6 = α * 42 => α = 6 / 42 = 1/7.

Now, we need another equation to solve for β. Since C* is a code in F, it must satisfy the conditions given in the problem statement. Specifically, since all codes in F have dimension 12 and contain the all-ones vector, their weight distributions are constrained accordingly.

But perhaps there's another known value of A_k(C*) that we can use. Alternatively, maybe we can find a relation using other known values from C1 and C2.

Wait, another approach: since any code in F is a linear combination of C1 and C2, the sum of coefficients α + β must account for the total number of vectors in C*. But I'm not sure if that's directly applicable here because each A_k is a separate equation.

Alternatively, perhaps we can use the fact that the all-ones vector is in C*, so there must be at least one vector of weight 24. However, this might not help us directly unless we know more about A_8(C*) or other weights.

Wait, another thought: since C* has dimension 12 and is a binary linear code, the total number of vectors in C* is 2^12 = 4096. So, the sum of all A_k from k=0 to 24 should equal 4096.

But that's a lot of equations. However, since we only need A_12(C*), maybe there's a smarter way.

Alternatively, perhaps we can use the fact that C2 has exactly 759 vectors of weight 8 (A_8(C2) = 759). So, if I can find another equation involving α and β using k=8, that would help.

So, let's consider A_8(C*). From the problem statement, we don't know A_8(C*) directly. But perhaps there is a relation or another known value.

Wait, maybe we can use the fact that C1 has a certain number of vectors of weight 8. Let me compute A_8(C1) from its weight enumerator polynomial.

From W_{C1}(x,y) = (x^8 + 14x^4y^4 + y^8)^3, we can find the coefficient of y^8 x^{24-8} = y^8 x^{16}.

To get a term with y^8, we need to consider all combinations where the exponents from each factor add up to 8. Let's see:

Each factor contributes either 0, 4, or 8 to the y exponent. So, possible ways to sum to 8 across three factors are:

• 8 + 0 + 0: pick y^8 once and x^8 twice.
• 4 + 4 + 0: pick two factors contributing y^4 and one contributing x^8.

Let's compute each case.

Case 1: Pick y^8 once, x^8 twice. The number of ways is C(3,1) = 3. Each such term contributes (y^8)^1 * (x^8)^2, so the coefficient from this case is 3*(1)^1*(14)^0 = 3.

Wait, no: actually, each factor can contribute either x^8, 14x^4y^4, or y^8. So, for Case 1, we pick one factor as y^8 and the other two as x^8. The coefficient is (1)^2 * (1) = 1 per selection, multiplied by C(3,1)=3, so total contribution is 3*1=3.

Case 2: Pick two factors contributing y^4 each and one factor contributing x^8. So, the number of ways is C(3,2)=3. Each such term contributes (14x^4y^4)^2 * x^8 = (14)^2 * x^{8+42} y^{42} = 196 x^{16} y^8.

Wait, but in terms of coefficients for y^8, each such term contributes 196. So, the total contribution from this case is 3 * 196 = 588.

Therefore, the coefficient of y^8 in W_{C1}(x,y) is 3 + 588 = 591. Hence, A_8(C1)=591.

Now, for C2, we know that A_8(C2)=759 and A_4(C2)=0.

So, if I can find another equation involving α and β using k=8, that would be helpful. However, the problem is that we don't know A_8(C*) directly. But perhaps there's a way to relate it through other known values or constraints.

Wait, maybe we can use the fact that C* has dimension 12, so the number of vectors in C* is 4096. Therefore, the sum over all k of A_k(C*) must be 4096.

But without knowing more specific A_k values, this might not help directly. Alternatively, perhaps we can use another known property or value from C1 and C2.

Alternatively, maybe there's a relation between the coefficients α and β based on other constraints, such as the presence of the all-ones vector.

Wait, since C* contains the all-ones vector, which has weight 24. So, A_24(C*) must be at least 1. But from the basis codes, perhaps we can find how many vectors of weight 24 are in C1 and C2.

Looking back at W_{C1}(x,y), the term with y^24 would come from picking y^8 from all three factors: (y^8)^3 = y^{24}. The coefficient is 1, so A_24(C1)=1.

For C2, since it's a code in F, it must also contain the all-ones vector. So, A_24(C2) must be at least 1. But I don't know if it has more vectors of weight 24. However, perhaps for simplicity, we can assume that both C1 and C2 have exactly one vector of weight 24 (the all-ones vector). Therefore, A_24(C*) = α1 + β1.

But since C* must contain the all-ones vector, A_24(C*) is at least 1. However, without more information, this might not help us directly unless we can find another equation.

Alternatively, perhaps there's a known relation between the coefficients α and β based on other properties of the codes. For example, maybe the sum of α and β must equal 1 because C* is a linear combination of C1 and C2. But I'm not entirely sure about that.

Wait, considering that any code in F can be expressed as a linear combination of C1 and C2, perhaps the coefficients α and β are such that α + β = 1. This would make sense if they form a basis, meaning that each code is uniquely represented by these two. So, assuming α + β = 1.

Given that we have α = 1/7 from earlier, then β = 1 - 1/7 = 6/7.

Now, with α and β known, we can compute A_12(C*) as:

A_12(C*) = α * A_12(C1) + β * A_12(C2)

But I don't know A_12(C1) or A_12(C2). So, I need to find these values.

First, let's compute A_12(C1). From W_{C1}(x,y), we can find the coefficient of y^12 x^{12} (since 24 - 12 = 12).

To get a term with y^12, we need to consider all combinations where the exponents from each factor add up to 12. Each factor contributes either 0, 4, or 8 to the y exponent.

Possible ways to sum to 12 across three factors:

• 8 + 4 + 0: pick one factor as y^8, another as y^4, and the third as x^8.
• 4 + 4 + 4: pick all three factors as y^4.

Let's compute each case.

Case 1: One factor contributes y^8, another contributes y^4, and the third contributes x^8. The number of ways is C(3,1)C(2,1) = 32=6 (choose which factor gives y^8, then which of the remaining two gives y^4). Each such term contributes (y^8)(y^4)(x^8) = y^{12} x^{8}. The coefficient for each selection is (1)(14)(1)=14. So, total contribution from this case is 6*14=84.

Case 2: All three factors contribute y^4. Each factor contributes 14x^4y^4. So, the product is (14)^3 x^{12} y^{12}. The coefficient is 14^3 = 2744. But wait, this would be if all three are contributing y^4, but each term in the expansion would have a coefficient of 14 from each factor. So, the total contribution is (14)^3 * C(3,3) = 2744*1=2744.

Wait, no: actually, when expanding (a + b + c)^n, each term's coefficient is determined by the multinomial coefficients. But in this case, since we're multiplying three identical factors, perhaps it's easier to think of it as:

The expansion of (x^8 + 14x^4y^4 + y^8)^3 will have terms where each factor contributes either x^8, 14x^4y^4, or y^8. So, for the case where all three contribute y^4, it's only one way (all three factors pick the middle term), and the coefficient is (14)^3 = 2744.

Therefore, total contribution from Case 2 is 2744.

Adding both cases: 84 + 2744 = 2828. So, A_12(C1)=2828.

Now, for C2, we need to find A_12(C2). But I don't have direct information about it. However, perhaps there's a way to relate it through other known values or properties.

Wait, considering that the weight distribution of any code in F is a linear combination of C1 and C2, and given that C* has exactly 6 vectors of weight 4, which comes from αA_4(C1) + βA_4(C2)=6. But we already used this to find α=1/7.

But perhaps there's another equation involving A_8 or other weights that can help us find A_12(C2).

Alternatively, maybe we can use the fact that the sum of all A_k for k=0 to 24 must equal 4096. But without knowing more specific values, this might not be directly helpful.

Wait, let's think about what we know:

• For C*, A_4(C*) = 6.
• From earlier, α = 1/7 and β = 6/7.

We need to find A_12(C*).

But to compute that, we need A_12(C1) and A_12(C2). We have A_12(C1)=2828. So, if we can find A_12(C2), we can proceed.

Alternatively, perhaps there's a way to express A_12(C*) in terms of other known quantities or through some combinatorial relations.

Wait, considering that C* is a code with certain properties, maybe the number of vectors of weight 12 can be determined using MacWilliams identities or other coding theory results. However, I'm not sure about the exact application here.

Alternatively, perhaps we can use the fact that the weight distribution must satisfy certain linear constraints based on the basis codes.

Wait, since any code in F is a linear combination of C1 and C2, their weight distributions are also linear combinations. Therefore, for each k, A_k(C*) = αA_k(C1) + βA_k(C2).

We already have:

A_4(C*) = 6 = α0 + β759 => Wait no, earlier I think there was a mistake.

Wait, let's go back. Earlier, we had:

For C*, A_4(C*)=6.

From the basis codes:

C1 has A_4(C1)=0 because W_{C1}(x,y) only has terms where y^k with k multiple of 8? Wait no, looking back at W_{C1}(x,y), it's (x^8 + 14x^4y^4 + y^8)^3. So, the exponents in y are multiples of 4: 0,4,8,... up to 24.

Wait, but for C1, A_4(C1) would be the coefficient of y^4 x^{20} in W_{C1}(x,y). Let's compute that.

To get a term with y^4, we need one factor contributing y^4 and the other two contributing x^8. So:

Number of ways: C(3,1)=3 (choose which factor contributes y^4).

Each such term has coefficient 14 from the middle term, and 1 from each of the other two factors.

So, total contribution is 3*14=42. Therefore, A_4(C1)=42.

Wait, earlier I thought that C1 had A_4=0, but that's incorrect. So, correcting that:

A_4(C1)=42.

Similarly, for C2, we have A_4(C2)=0 and A_8(C2)=759.

So, going back to the equation for C*:

A_4(C*) = αA_4(C1) + βA_4(C2) => 6 = α42 + β0 => 6=42α => α=6/42=1/7. Which matches our earlier result.

Now, to find A_8(C*), we can use:

A_8(C*) = αA_8(C1) + βA_8(C2)

We have A_8(C1)=591 and A_8(C2)=759.

So,

A_8(C*)= (1/7)*591 + (6/7)*759

Let's compute this:

(591)/7 = 84.42857...

(6*759)/7 = (4554)/7 ≈ 650.5714...

Adding together: 84.42857 + 650.5714 ≈ 735.

But let's compute it exactly:

591 ÷ 7 = 84 with remainder 3 (since 7*84=588, 591-588=3). So, 591/7 = 84 + 3/7.

Similarly, 6759 = 4554. 4554 ÷7: 7650=4550, remainder 4. So, 4554/7=650 +4/7.

Adding both fractions:

3/7 +4/7=7/7=1.

So total A_8(C*)=84+650 +1=735.

Now, perhaps we can use the fact that the sum of all A_k for k=0 to 24 is 4096. But without knowing more specific values, this might not help directly.

Alternatively, maybe there's a way to express A_12(C*) in terms of other known quantities or through some combinatorial relations.

Wait, considering that the weight distribution must satisfy certain linear constraints based on the basis codes, perhaps we can find A_12(C2) by using another equation involving known values.

But without more information about C2's weight distribution beyond A_4=0 and A_8=759, it's challenging to proceed directly. However, perhaps there's a way to relate A_12(C*) through other means.

Alternatively, maybe we can use the fact that the number of vectors in C* is 4096, so sum_{k} A_k =4096. But without knowing more specific values, this might not help directly.

Wait, perhaps there's a symmetry or duality in the weight distribution due to the properties of the code. For example, in some codes, the number of vectors of weight k is equal to the number of vectors of weight 24 -k. However, I'm not sure if that applies here.

Alternatively, considering that C* contains the all-ones vector (weight 24), perhaps there's a relation between A_k and A_{24-k}.

But without more information, it's hard to apply this directly.

Wait, another approach: since we have two basis codes, C1 and C2, their weight distributions form a basis for the space of possible weight distributions. Therefore, any code in F can be expressed as αC1 + βC2, where α and β are scalars such that α+β=1 (since it's a linear combination in terms of coefficients adding to 1).

Given that, perhaps we can express A_12(C*) = αA_12(C1) + βA_12(C2). We already have α=1/7 and β=6/7.

We need to find A_12(C2). But how?

Wait, perhaps there's another equation involving A_8 or other weights that can help us find A_12(C2).

Alternatively, considering that the sum of all A_k for C2 must be 4096, and we know some specific values:

For C2: A_0=1 (since it's a linear code containing the zero vector) A_4=0 A_8=759 We need to find other A_k's.

But without more information, this is difficult. However, perhaps there's a way to use the fact that the weight distribution must satisfy certain properties due to the code being self-dual or having certain symmetries.

Alternatively, considering that C2 has no vectors of weight 4 and exactly 759 vectors of weight 8, maybe we can find A_12(C2) by using some combinatorial relations or known results about binary codes with specific properties.

Wait, perhaps the code C2 is related to the extended binary Golay code. The extended binary Golay code has parameters [24,12,8], and its weight distribution includes 759 vectors of weight 8. This matches the description of C2: dimension 12, minimum distance 8 (since A_4=0), and A_8=759.

Therefore, C2 is likely the extended binary Golay code. If that's the case, then we can look up its weight distribution to find A_12(C2).

Looking up the weight distribution of the extended binary Golay code (which has parameters [24,12,8]), we find:

The number of codewords of each weight is as follows:

• Weight 0: 1
• Weight 8: 759
• Weight 12: 2576
• Weight 16: 759
• Weight 24: 1

So, A_12(C2)=2576.

Now, we can compute A_12(C*)= αA_12(C1) + βA_12(C2)

We have:

α=1/7, β=6/7 A_12(C1)=2828 A_12(C2)=2576

So,

A_12(C*) = (1/7)*2828 + (6/7)*2576

Compute each term:

(1/7)*2828 = 404

(6/7)2576: Let's compute 2576 ÷7=368, so 6368=2208.

Adding together: 404 +2208=2612.

Therefore, the number of vectors in C^* with weight 12 is 2612. </think>

[THOUGHT] To solve this problem, we need to determine the number of vectors of weight 12 in a specific code ( C^* ) within the family ( \mathcal{F} ). The code ( C^* ) has exactly 6 vectors of weight 4 and is expressed as a linear combination of two basis codes ( C_1 ) and ( C_2 ).

First, we identify that ( C_1 ) has a known weight enumerator polynomial, which allows us to compute its weight distribution. Specifically, the number of vectors of weight 12 in ( C_1 ) is found by expanding the given polynomial.

Next, recognizing that ( C_2 ) corresponds to the extended binary Golay code, we can look up its weight distribution. This code has exactly 2576 vectors of weight 12.

Given that any code in ( \mathcal{F} ) is a linear combination of ( C_1 ) and ( C_2 ), we express the weight distribution of ( C^* ) as: [ A_{12}(C^*) = \alpha \cdot A_{12}(C_1) + \beta \cdot A_{12}(C_2) ] where ( \alpha = \frac{1}{7} ) and ( \beta = \frac{6}{7} ).

Substituting the known values: [ A_{12}(C^*) = \left(\frac{1}{7}\right) \cdot 2828 + \left(\frac{6}{7}\right) \cdot 2576 ] Calculating each term and summing them gives the final result. [/THOUGHT]

[PYTHON]

Calculate A_12(C^*) using the given coefficients

alpha = 1 / 7 beta = 6 / 7

A_12_C1 = 2828 A_12_C2 = 2576

A_12_C_star = alpha * A_12_C1 + beta * A_12_C2 print(int(A_12_C_star))

<think> Okay, let's see what I need to do here. The problem is about coding theory, specifically dealing with binary linear codes of length 24 and dimension 12. The family F consists of all such codes where every vector has a Hamming weight that's a multiple of 4, and the all-ones vector is included.

We're given two basis codes, C₁ and C₂. For C₁, its weight enumerator polynomial is provided as (x⁸ + 14x⁴y⁴ + y⁸)³. I remember that expanding this polynomial will give us the coefficients A_k(C₁), which represent the number of vectors of weight k in C₁.

For C₂, it's mentioned that there are no vectors of weight 4 and exactly 759 vectors of weight 8. This suggests that C₂ is a specific code with known properties, possibly related to the extended binary Golay code, which I think has parameters [24, 12, 8] and includes 759 codewords of weight 8.

The problem states that any code in F can be expressed as a linear combination of C₁ and C₂. So, for our specific code C*, which has exactly 6 vectors of weight 4, we need to find the number of vectors of weight 12.

First, I'll calculate A₄(C₁) by expanding its polynomial. The term x⁸ + 14x⁴y⁴ + y⁸ is symmetric and when raised to the third power, it will give us a polynomial where each term corresponds to a specific weight distribution. Expanding this should give me the coefficients for A₄(C₁) and A₈(C₁).

Next, since C₂ has no vectors of weight 4 (A₄(C₂)=0) and exactly 759 vectors of weight 8 (A₈(C₂)=759), I can set up a system of equations to solve for the coefficients α and β in the linear combination that expresses C* as αC₁ + βC₂.

Given that A₄(C*) = 6, which is equal to αA₄(C₁) + βA₄(C₂). Since A₄(C₂)=0, this simplifies to αA₄(C₁) = 6. Similarly, for the weight 8 vectors, we have A₈(C) = αA₈(C₁) + β759.

Once I find α and β, I can then compute A₁₂(C*) using these coefficients with the known values of A₁₂(C₁) and A₁₂(C₂).

After calculating all this, I'll use Python to verify my computations and ensure that there are no errors in the arithmetic or polynomial expansion. This will give me the final number of vectors of weight 12 in C*. </think>

The number of vectors in ( C^* ) with Hamming weight 12 is:

\boxed{2612}