Weight distribution of hexacode

Question

I am given the generator matrix of the hexacode over $\mathbb{F}_2(\omega)/(\omega^2+\omega+1)=\{0,1,\omega, 1+\omega \}$ as $$ G=\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & \omega & \omega \\ 0 & 1 & 0 & \omega & 1 & \omega \\ 0 & 0 & 1 & \omega & \omega & 1 \end{array} \right]. $$ I am asked to calculate the minimum distance $d$ of the code and the weight distribution. I don't see any way to answer this short of writing out all $64$ codewords and then inspecting the list to determine my answers. Is there a way to analyze this a little more intelligently?

I know that the code is Hermitian self-dual and so the codewords all have even weight. That tells me $d\ge 2$ and since the rows of $G$ all have $4$ nonzero entries, $d\le 4$. I suspect that $d=4$ and that the weight distribution is $A_0=1,~A_2=0,~A_4=a,~\text{and}~A_6=b$, but I can't see an argument on why there are no codewords of weight $2$ or how the weights of the nonzero codewords are distributed.

UPDATE: After discussing this further with my professor, it seems the only real way to tackle this is brute force. To solve, you are basically left to writing out all 64 codewords and inspecting the list.

Answer 1

You can do this using a computational package relatively easily.

For example, using the Magma Online Calculator you can set up the Hexacode $C$ (note that there is no ambiguity regarding omega because there is only one quadratic irreducible polynomial over $\mathbb{F}_{2}$):

F := GF(4);
V := VectorSpace(F, 6);
G := {V![1,0,0,1,omega,omega], V![0,1,0,omega,1,omega], V![0,0,1,omega,omega,1]};
C := sub;

And then compute and print your weight distribution:

WeightDistribution := {* Weight(w) : w in C *};
print WeightDistribution;

(Magma has some nice sophisticated tools for working with codes, but this is just a quick naive way to form the vectors and compute their weights)