Given a $[5,3,3]_4$ perfect linear code, how is it possible to work out the number of cosets of this code with coset leader of given weight $i$?
I've tried messing about with the hamming bound and found that the number of distinct cosets is $\frac{4^5}{64} = 16$, but I can't seem to get any further.
There can only be one coset with coset leader of weight $0$, but how do I work out the rest?
Note that the number of distinct cosets of a linear code is given by $q^{n-k}$, thus there are $4^{5-3}=16$ cosets. The first coset of course will just be the code itself (trivial coset). To get the other cosets, pick an element of $F_4^5$ which is not yet listed as an element of any coset, and employ vector addition for that particular element to every element of the trivial coset. The resulting vectors will then form a new coset. You can easily check the coset leader since it should have the least (Hamming) weight amongst the vectors in a coset (NOTE: sometimes it might not be unique). Repeat the process until you get all 16 cosets (and coset leaders).