I need to find $r,q$ so that $Ham(r,q)$ is selfdual code and then $r,q$ so that it is a MDS code
My effort
$Ham(r,q)= [\frac{q^{r}-1}{q-1}, \frac{q^{r}-1}{q-1} -r, 3] $ code
A code $C$ of length $n$ is selfdual iff $dim C= \frac{n}{2}$
So in this case $\frac{q^{r}-1}{q-1} -r = \frac{\frac{q^{r}-1}{q-1}}{2} $ and this gives me $\frac{1}{2} \frac{q^{r}-1}{q-1} = r $
A linear code $[n,k,d]$ is called MDS if $k+d=n+1$
So in this case $\frac{q^{r}-1}{q-1} -r + 3 = \frac{q^{r}-1}{q-1} +1$, so $r=2$
Is there anything else I can do?
$Ham(r,q)= [\frac{q^{r}-1}{q-1}, \frac{q^{r}-1}{q-1} -r, 3] $
A code is self-dual iff it's equal to its dual. $S(r,q)= [\frac{q^{r}-1}{q-1}, r, q^{r-1}]$ , so $q^{r-1} = 3$ and that gives us $q=3, r=2$ , the only correct pair.
A linear code $[n,k,d]$ is called MDS if $k+d=n+1$
So in this case $\frac{q^{r}-1}{q-1} -r + 3 = \frac{q^{r}-1}{q-1} +1$, so $r=2$ Finally, $Ham(2,q)= [q+1, q-1, 3]$ is a MDS code.