Problem Analysis #8

[THOUGHT] We are given that any code C $$\in$$ $$\mathcal{F}$$ has a weight distribution that is a linear combination of the weight distributions of two basis codes $C_1$ and $C_2$. We are told:

• C^* $$\in$$ $$\mathcal{F}$$, so its weight distribution is A_k(C^*) = a $$\cdot$$ A_k(C_1) + b $$\cdot$$ A_k(C_2) for some scalars $a, b$.
• $C^*$ has exactly 6 vectors of weight 4: $A_4(C^*) = 6$.
• $C_1$: weight enumerator $W_{C_1}(x,y) = (x^8 + 14x^4y^4 + y^8)^3$
• $C_2$: $A_4(C_2) = 0$, $A_8(C_2) = 759$

We are to compute $A_{12}(C^*)$, the number of weight-12 vectors in $C^*$.

Step 1: Expand $W_{C_1}(x,y) = (x^8 + 14x^4y^4 + y^8)^3$ to get $A_k(C_1)$ for $k = 4, 8, 12$, etc.

Note: $x^{24 - \text{wt}(v)} y^{\text{wt}(v)}$, so the exponent of $y$ is the weight.

We need to compute the coefficient of $y^4$, $y^8$, and $y^{12}$ in $W_{C_1}(x,y)$.

Let’s expand $(x^8 + 14x^4y^4 + y^8)^3$.

Let $a = x^8$, $b = 14x^4y^4$, $c = y^8$

We are to expand $(a + b + c)^3$, and collect terms where the total power of $y$ is 4, 8, 12.

• $y^4$: only from $b$, and $b$ has $y^4$, so we need exactly one $b$, and the rest $a$ or $c$, but $c = y^8$, so $y^8$ would overshoot.

So $y^4$: only terms with exactly one $b$, and two $a$'s.

• $b$: $14x^4y^4$
• $a$: $x^8$

So term: \binom{3}{1} a^2 b = 3 $$\cdot$$ (x^8)^2 $$\cdot$$ (14x^4y^4) = 3 $$\cdot$$ x^{16} $$\cdot$$ 14x^4y^4 = 42 x^{20} y^4

So $A_4(C_1) = 42$

• $y^8$: can come from:
  • Two $b$'s: $b^2 = (14x^4y^4)^2 = 196 x^8 y^8$, and one $a$: $x^8$
    • \binom{3}{2} a b^2 = 3 $$\cdot$$ x^8 $$\cdot$$ 196 x^8 y^8 = 588 x^{16} y^8
  • One $c$: $y^8$, and two $a$'s: $x^{16}$
    • \binom{3}{1} a^2 c = 3 $$\cdot$$ x^{16} $$\cdot$$ y^8 = 3 x^{16} y^8
  • Other combinations? $b$ and $c$: b $$\cdot$$ c = 14x^4y^4 $$\cdot$$ y^8 = 14x^4 y^{12} → too high
  • $b^2$: already done
  • $c$: only one $c$, and $a^2$: done

So total $y^8$: $588 + 3 = 591$, so $A_8(C_1) = 591$

• $y^{12}$: terms with $y^{12}$
  • Three $b$'s: $b^3 = (14x^4y^4)^3 = 2744 x^{12} y^{12}$, coefficient $\binom{3}{3} = 1$ → $2744 x^{12} y^{12}$
  • One $b$ and one $c$: b $$\cdot$$ c = 14x^4y^4 $$\cdot$$ y^8 = 14x^4 y^{12}, and one $a$
    • $\binom{3}{1,1,1} = 6$ ways: choose which is $b$, which is $c$, which is $a$
    • 6 $$\cdot$$ x^8 $$\cdot$$ 14x^4y^4 $$\cdot$$ y^8 = 6 $$\cdot$$ 14 x^{12} y^{12} = 84 x^{12} y^{12}
  • Two $c$'s: $c^2 = y^{16}$, too high
  • So only two cases: $b^3$ and $a b c$

Thus $A_{12}(C_1) = 2744 + 84 = 2828$

Now, for $C_2$:

• $A_4(C_2) = 0$
• $A_8(C_2) = 759$
• We don’t know $A_{12}(C_2)$, but we can find it from the fact that $C_2$ is a 12-dimensional binary linear code of length 24, with all weights divisible by 4, and $$\mathbf{1}$$ $$\in$$ C_2. So $A_0 = 1$, $A_{24} = 1$, and $A_k = 0$ unless k $$\equiv$$ 0 \pmod{4}.

Also, since $C_2$ is in $$\mathcal{F}$$, its weight distribution is a linear combination of $C_1$ and $C_2$, but $C_2$ is a basis, so we can use the known values.

But we are not given $A_{12}(C_2)$, but we can use the fact that $C^* = a C_1 + b C_2$, and we know $A_4(C^*) = 6$

We have:

• A_4(C^*) = a $$\cdot$$ A_4(C_1) + b $$\cdot$$ A_4(C_2) = a $$\cdot$$ 42 + b $$\cdot$$ 0 = 42a
• Set equal to 6: 42a = 6 $$\Rightarrow$$ a = 6 / 42 = 1/7

Now, A_{12}(C^*) = a $$\cdot$$ A_{12}(C_1) + b $$\cdot$$ A_{12}(C_2)

We know $a = 1/7$, $A_{12}(C_1) = 2828$

But we don’t know $b$, nor $A_{12}(C_2)$

But we can use another known value: $C_2$ has $A_8(C_2) = 759$

So A_8(C^*) = a $$\cdot$$ A_8(C_1) + b $$\cdot$$ A_8(C_2) = (1/7)(591) + b $$\cdot$$ 759

But we don’t know $A_8(C^*)$, so we need another way.

Wait — but $C^*$ is in $$\mathcal{F}$$, so it must satisfy:

• $\dim C^* = 12$, so $|C^*| = 2^{12} = 4096$
• $A_0 = 1$, $A_{24} = 1$, and $A_k = 0$ unless k $$\equiv$$ 0 \pmod{4}
• $A_4(C^*) = 6$
• We are to compute $A_{12}(C^*)$

We can use the fact that $C^* = a C_1 + b C_2$, and $a = 1/7$, so C^* = $$\frac{1}{7}$$ C_1 + b C_2

But $b$ is not known.

But since $C^*$ must be a linear code, and $C_1, C_2$ are basis codes, and $C^*$ is in $$\mathcal{F}$$, the coefficients $a, b$ must be such that $C^*$ has integer weight counts.

But we can use the total number of vectors: $$\sum_{k=0}^{24}$$ A_k = 4096

But we don’t know $A_k$ for all $k$, but we can use the known values.

Alternatively, note that $C_2$ is a known code: it is likely the extended binary Golay code, but wait — the extended Golay code has $A_0 = 1, A_8 = 759, A_{12} = 2576, A_{16} = 759, A_{24} = 1$, and $A_4 = 0$, $A_{20} = 0$, etc.

In fact, the extended binary Golay code $$\mathcal{G}_{24}$$ is a $[24,12,8]$ code with $A_8 = 759$, $A_{12} = 2576$, $A_{16} = 759$, $A_0 = 1$, $A_{24} = 1$, and $A_4 = 0$, $A_{20} = 0$, and $A_k = 0$ for k \not$$\equiv$$ 0 \pmod{4}.

So $C_2$ is the extended Golay code.

Similarly, $C_1$: $(x^8 + 14x^4y^4 + y^8)^3$

Let’s compute $A_0(C_1)$: when $y^0$, so $x^{24}$, only from $x^8$ terms.

Each factor contributes $x^8$, so $(x^8)^3 = x^{24}$, so $A_0 = 1$

Similarly, $A_{24} = 1$, from $y^8$ terms.

Now, $C^* = a C_1 + b C_2$, $a = 1/7$

But $C^*$ must have integer $A_k$, so $b$ must be chosen so that $A_k(C^*)$ are integers.

But we can use the fact that $C^*$ has $A_4 = 6$, and $A_4(C_1) = 42$, $A_4(C_2) = 0$, so $a = 6/42 = 1/7$

Now, A_{12}(C^*) = (1/7) $$\cdot$$ 2828 + b $$\cdot$$ A_{12}(C_2)

But $A_{12}(C_2) = 2576$ (known from extended Golay code)

So A_{12}(C^*) = 2828 / 7 + b $$\cdot$$ 2576

$2828 / 7 = 404$

So $A_{12}(C^*) = 404 + 2576 b$

Now, $C^*$ must have total size 4096, and $A_0 = 1$, $A_{24} = 1$, and $A_k = 0$ for $k$ not divisible by 4.

But we can use another known value: $C^*$ must contain $$\mathbf{1}$$, so $A_{24} = 1$, $A_0 = 1$, and $A_{12}$ is symmetric.

But we need to determine $b$. But $b$ is not known.

Wait — but $C^*$ is a linear code, so $A_k$ must be integers, and $b$ must be rational.

But $C^*$ is in $$\mathcal{F}$$, so it must be a linear combination with rational coefficients, but since $C_1$ and $C_2$ are basis, and $C^*$ is in $$\mathcal{F}$$, it must be that $b$ is rational.

But we can use the fact that $C^*$ has $A_4 = 6$, and $A_8$ must be determined.

But we don't know $A_8(C^*)$, but we can use the fact that $C^*$ must have A_8(C^*) = (1/7) $$\cdot$$ 591 + b $$\cdot$$ 759

$591 / 7 = 84.428...$ — not integer!

But $A_8(C^*)$ must be integer.

Contradiction?

Wait — $A_8(C_1) = 591$, $a = 1/7$, so a $$\cdot$$ A_8(C_1) = 591 / 7 = 84.428... — not integer.

But $A_8(C^*)$ must be integer.

So unless $b$ compensates, but $b$ is multiplied by 759, so $b$ must be rational.

But $591 / 7 = 84.428...$, not integer.

But $A_8(C^*)$ must be integer, so (1/7)(591) + b $$\cdot$$ 759 must be integer.

But $591 / 7 = 84.428...$, not integer.

But $591 \div 7 = 84.428...$ — 7*84 = 588, 591 - 588 = 3, so $591/7 = 84 + 3/7$

So $A_8(C^*) = 84 + 3/7 + 759 b$

Set $b = m/n$, but we need 84 + 3/7 + 759 b $$\in$$ $$\mathbb{Z}$$

So $759 b = k - 84 - 3/7$, so $b = (k - 84 - 3/7) / 759$

But $b$ must be such that $A_{12}(C^*) = 404 + 2576 b$ is integer.

But we have a problem: $A_4(C^*) = 6$, $A_4(C_1) = 42$, $A_4(C_2) = 0$, so $a = 6/42 = 1/7$, so $a$ is fixed.

But A_8(C^*) = (1/7)(591) + b $$\cdot$$ 759 = 591/7 + 759 b

$591/7 = 84.428...$, not integer.

But $A_8(C^*)$ must be integer.

So unless $b$ is chosen to make it integer, but $b$ is not free — $C^*$ is determined by $A_4 = 6$, so $a = 1/7$, but $b$ is not determined yet.

But $C^*$ is in $$\mathcal{F}$$, so it must be a linear combination, but the coefficients must be such that all $A_k$ are integers.

But $C_1$ and $C_2$ are basis codes, so any C $$\in$$ $$\mathcal{F}$$ is $a C_1 + b C_2$, but $a, b$ must be rational.

But $a = 1/7$, so $C^* = (1/7) C_1 + b C_2$

Now, $A_8(C^*) = 591/7 + 759 b = 84 + 3/7 + 759 b$

Set this to be integer: 84 + 3/7 + 759 b = N $$\in$$ $$\mathbb{Z}$$

So $759 b = N - 84 - 3/7 = (7(N - 84) - 3)/7$

So b = $$\frac{7(N - 84) - 3}{7 \cdot 759}$$

Must be rational, but we need $A_{12}(C^*) = 404 + 2576 b$ to be integer.

But we have another condition: $C^*$ must contain $$\mathbf{1}$$, so $A_{24} = 1$, which is satisfied since both $C_1$ and $C_2$ have $A_{24} = 1$, so $A_{24}(C^*) = (1/7)(1) + b(1) = 1/7 + b$

But $A_{24}(C^*) = 1$, since $$\mathbf{1}$$ $$\in$$ C^*

So 1/7 + b = 1 $$\Rightarrow$$ b = 1 - 1/7 = 6/7

Oh! We forgot that $C^*$ contains $$\mathbf{1}$$, so $A_{24}(C^*) = 1$

But A_{24}(C^*) = a $$\cdot$$ A_{24}(C_1) + b $$\cdot$$ A_{24}(C_2) = a $$\cdot$$ 1 + b $$\cdot$$ 1 = a + b

Set equal to 1: $a + b = 1$

We have $a = 1/7$, so $b = 1 - 1/7 = 6/7$

Now we can compute A_{12}(C^*) = a $$\cdot$$ A_{12}(C_1) + b $$\cdot$$ A_{12}(C_2) = (1/7)(2828) + (6/7)(2576)

Compute:

• $2828 / 7 = 404$
• 6 $$\cdot$$ 2576 = 15456
• $15456 / 7 = 2208$

So $A_{12}(C^*) = 404 + 2208 = 2612$

Now verify A_8(C^*) = (1/7)(591) + (6/7)(759) = (591 + 6 $$\cdot$$ 759)/7

Compute 6 $$\cdot$$ 759 = 4554

$591 + 4554 = 5145$

$5145 / 7 = 735$

Integer, good.

$A_4(C^*) = (1/7)(42) + (6/7)(0) = 6$, correct.

$A_0 = (1/7)(1) + (6/7)(1) = 1$, good.

$A_{24} = 1$, good.

Now, total sum: $$\sum$$ A_k = 4096

We can check if $A_{12} = 2612$ makes sense.

But we have $A_{12} = 2612$, and $A_8 = 735$, $A_4 = 6$, $A_0 = 1$, $A_{24} = 1$

Also $A_{16} = a A_{16}(C_1) + b A_{16}(C_2)$

$A_{16}(C_1)$: from earlier, $y^{16}$, so $b^2 c$ or $a c^2$ or $c^2 a$, etc.

From $(x^8 + 14x^4y^4 + y^8)^3$

$y^{16}$: from $c^2$ and $a$: $c^2 = y^{16}$, $a = x^8$, so $\binom{3}{1} a c^2 = 3 x^8 y^{16}$

$b^2$: $(14x^4y^4)^2 = 196 x^8 y^8$, times $c = y^8$, so $y^{16}$, and $a$: but $b^2 c$: $b^2$ gives $y^8$, $c$ gives $y^8$, so $y^{16}$, and $a$: so $b^2 c$: $\binom{3}{2,1} = 3$ ways, (14x^4y^4)^2 $$\cdot$$ y^8 = 196 $$\cdot$$ 14 x^8 y^{16}? No

$b^2$: $(14x^4y^4)^2 = 196 x^8 y^8$, then times $c = y^8$, so $y^{16}$, and $x^8$, so 196 $$\cdot$$ 1 $$\cdot$$ x^8 y^{16}, and coefficient $\binom{3}{2,1} = 3$, so 3 $$\cdot$$ 196 = 588

$c^2$: $(y^8)^2 = y^{16}$, and $a = x^8$, so $\binom{3}{1} = 3$ ways, 3 $$\cdot$$ 1 $$\cdot$$ x^8 y^{16}, so $3$

$b^4$ not possible.

So $A_{16}(C_1) = 588 + 3 = 591$

$A_{16}(C_2) = 759$ (since extended Golay code has $A_{16} = 759$)

So $A_{16}(C^*) = (1/7)(591) + (6/7)(759) = (591 + 4554)/7 = 5145/7 = 735$

Same as $A_8$, which makes sense by symmetry.

Now $A_{12} = 2612$, as computed.

So the answer is 2612.

But let me confirm $A_{12}(C_1)$: earlier I said $2828$, but let me double-check.

$C_1$: $(x^8 + 14x^4y^4 + y^8)^3$

$y^{12}$: from $b^3$: $(14x^4y^4)^3 = 14^3 x^{12} y^{12} = 2744 x^{12} y^{12}$, and $\binom{3}{3} = 1$, so $2744$

$b^2 c$: (14x^4y^4)^2 $$\cdot$$ y^8 = 196 x^8 y^8 $$\cdot$$ y^8 = 196 x^8 y^{16}, not $y^{12}$

$b c^2$: (14x^4y^4) $$\cdot$$ (y^8)^2 = 14 x^4 y^4 $$\cdot$$ y^{16} = 14 x^4 y^{20}, not $y^{12}$

$a b c$: x^8 $$\cdot$$ 14x^4y^4 $$\cdot$$ y^8 = 14 x^{12} y^{12}, and $\binom{3}{1,1,1} = 6$, so 6 $$\cdot$$ 14 = 84

$a^2 c$: (x^8)^2 $$\cdot$$ y^8 = x^{16} y^8, not $y^{12}$

$a c^2$: x^8 $$\cdot$$ y^{16}, not $y^{12}$

$b^3$: $14^3 = 2744$, as above

$a b c$: 6 $$\cdot$$ 14 = 84

$c^3$: $y^{24}$, not $y^{12}$

$a^3$: $x^{24}$, not $y^{12}$

$b^2 c$: $y^8$ from $b^2$, $y^8$ from $c$, so $y^{16}$, not $y^{12}$

So only $b^3$ and $a b c$

$b^3$: $14^3 = 2744$

$a b c$: 6 $$\cdot$$ 14 = 84

Total $2744 + 84 = 2828$, yes.

$A_{12}(C_2) = 2576$, known.

So A_{12}(C^*) = (1/7)(2828) + (6/7)(2576) = 404 + (6 $$\cdot$$ 2576)/7

6 $$\cdot$$ 2576 = 15456

$15456 / 7 = 2208$

$404 + 2208 = 2612$

Yes.

So the number of codewords of weight 12 in $C^*$ is 2612.

Final Answer

$$\boxed{2612}$$