I've just started reading a book on Error-Correcting Codes; it's my first contact with this subject. I'm stuck on the following exercise:
Given the alphabet $A=\{a,b,c,d\}$ and codes $C_1=\{aac,acd,cca,acd,aaa\}$, $C_2=\{abc,bbb,ddb,dda,dbc\}$, and $C_3=\{aac,add,cba,cca,aaa\}$, identify pairs of equivalent codes.
What I've tried so far is:
I've noticed that acd appears in $C_1$ twice, I don't know if that's a typo, but if it isn't then $C_1$ cannot be equivalent to any of the other codes, because no two elements in either $C_2$ or $C_3$ are the same. To find whether $C_2$ and $C_3$ are equivalent or not, I've made the following tables:
Hamming Distance of elements in $C_2$: \begin{array}{llllll} d(u,v) & abc & bbb & ddb & dda & dbc\\ abc & 0 & 2 & 3 & 3 & 1 \\ bbb & 2 & 0 & 2 & 3 & 2 \\ ddb & 3 & 2 & 0 & 1 & 2 \\ dda & 3 & 3 & 1 & 0 & 2 \\ dbc & 1 & 2 & 2 & 2 & 0 \end{array}
Hamming Distance of elements in $C_3$: \begin{array}{llllll} d(u,v) & aac & add & cba & cca & aaa\\ aac & 0 & 2 & 3 & 3 & 1 \\ add & 2 & 0 & 3 & 3 & 2 \\ cba & 3 & 3 & 0 & 1 & 2 \\ cca & 3 & 3 & 1 & 0 & 2 \\ aaa & 1 & 2 & 2 & 2 & 0 \end{array}
Because these two tables are not equivalent, due to $d(ddb, bbb)=2 \neq 3=d(cba,add)$, I believe the codes also are not.
Are there really no pairs of equivalent codes in this exercise? Are there better ways of showing that two codes are equivalent, other than looking at a table of distances or just guessing the right isomorphism?