Minimum distance of a ternary linear code

Question

How large can the minimum distance of a ternary linear code of length $n$ and dimension $k=n^{0.99}$ be?

Clearly, it can be $n^{0.01}$: by choosing basis vectors with exactly $n^{0.01}$ many nonzero coordinates, in such a way that these blocks are pairwise disjoint. Can it be larger? If yes, how large can it be? Maybe, it's trivial, but I don't see the answer right now.

Answer 1

There exists nontrivial ternary MDS codes, so $d = n-k+1$ is possible for some choice of parameters. See the Singleton bound. It's far from certain that one exists for $k = n^{.99}$ however because it would require that by some voodoo a ternary MDS code would exist for a very specific set of parameters. It's easier to construct ternary MDS codes than binary, but not THAT much easier...

If you're fine with $k = \lfloor n^{.99} \rfloor$ or similar, then it's very likely $d = n-k+1$ is possible, because you could probably construct a single parity check code for these parameters (the single parity check code is MDS).

Answer 2

Thanks for your answer, and sorry for the slow replying, I was travelling and then I haven't noticed the answer.

So it can be so close to the Singleton-bound? ($k\approx n^{0.99}$ is fine)

I would be really interested in an easy example when the minimum distance is at least, say $n^{0.99}$ (which is still far from the Singleton-bound).