I'm not really good at doing this type of exercices. But I'd like to know how to prove that ther are 132 words of weight 5 in the Ternary Golay Code. I am not allowed to use the weight enumerator.

I tried to ask the same question in the global code $GF(3)^{11}$ but not succeeded. So I'm quite suck on this.

Any suggestion would be aweomse.

  • 1
    $\begingroup$ Often a useful approach is to count the number of some objects in two different ways, possibly in terms of some unknown quantities, equating the results, and then solving for one of the previously unknown quantities. $\endgroup$ – Jyrki Lahtonen Dec 4 '18 at 13:07
  • $\begingroup$ You are talking only about Coding Theory or in general? $\endgroup$ – Lecter Dec 4 '18 at 16:01

I assume that you are expected to answer this question using only the (big) piece of information that the ternary Golay code $G$ is a perfect code with covering radius $\rho=2$.

An attack (filling in the details as the OP solved the problem themself):

  • The number of vectors of weight three in the space $GF(3)^{11}$ is equal to $\binom{11}3\cdot2^3=1320$. We can choose the three non-zero positions in $\binom{11}3$ ways, and each of those non-zero positions can have either $1$ or $2$ as the entry, so those three non-zero positions can be filled in $2^3$ different ways.
  • The covering property implies that a vector $x$ of weight is within Hamming distance $\le 2$ of a unique word $w\in G$. The triangle inequality implies that the weight of $w$ must be in the interval $[1,5]$. The minimum distance of $G$ is five, so $w$ must be of weight five exactly.
  • Given a codeword $w\in G$ of weight five, it is at distance two from exactly $\binom52=10$ vectors $x$ of weight three. This is because we get all such vectors $x$ by replaing two of the five non-zero components of $w$ with a zero. Note that the answer is independent of the choice of $w$.
  • Let the number of codewords of weight five be $M$. In light of the previous bullet, between them they cover $10M$ vectors of weight three. Observe that there is no overlap, for if two distinct codewords were both within distance two of the same vector $x$, then the distance between them would be at most four in violation of the known minimum distance of $G$.
  • Combining the first and the fourth bullet, we arrive at the equation $1320=10M$ implying $M=132$.
  • $\begingroup$ a) $8\binom{11}{3}$ b) Because it's a perfect code c) $10$ d)$10M$ e)Just solve $10M=8\binom{11}{3}$ $\endgroup$ – Lecter Dec 4 '18 at 15:41
  • $\begingroup$ Correct! Good job, @Lecter! $\endgroup$ – Jyrki Lahtonen Dec 4 '18 at 17:52

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