What is the minimal Hamming distance of concatenation of some word and a part of it encoded using Hamming codes?

Question

A word $M$ is of $n^2$ bits, $n>3$ is arranged in a $n\times n$ matrix. $A$ is the main diagonal of the matrix (that is elements $a_{i,j}$, $i=j$) and is encoded with Hamming code, resulting in $B$. The final result transmitted is word $M$ concatenated with $B$ (let us call it $R$). What is the minimal Hamming distance of $R$?

The minimal Hamming distance of $B$ is $3$ as holds with any Hamming code. $M$ is not encoded at all so its Hamming distance is $1$. Although $B$ is part of $R$ but $M$ is also part of $R$ and we have no way of knowing if a potential error comes from $B$ or $M$ so the minimal Hamming distance of $R$ is $1$ as well.

Am I in the right direction?

Answer 1

That is correct. The off-diagonal bits are not protected at all in this scheme, and toggling one of them does not affect any other bits. Therefore the minimum distance is $1$.

Answer 2

Yes, your intuition is right.

Since any distinct $M_1$ and $M_2$ must differ in one bit in the $M$ part, we know that the minimum Hamming distance $d_{\min}$ of this concatenated code must be greater than $1$. Now, consider two words $M_1$ and $M_2$ that differs in one off-diagonal position. Clearly, the Hamming distance between the corresponding (concatenated) codewords is $1$, which achieves the lower bound and hence $d_{\min}=1$.