Given data word $M$ of $2n$ bits $M$ is split into 2 words $M_1M_2$ whereas the length of $M_1$ is $n$ bits as well as the length of $M_2$ is also $n$ bits. $M_1$ and $M_2$ are encoded separately using Hamming code and result in words $K_1$ and $K_2$ respectively. The final result is the concatenation of $K_1$ and $K_2$ that is $K_1K_2$. What is the minimal Hamming distance of the code?
In general a code is said to be $k$ error detecting if, and only if, the minimum Hamming distance between any two of its codewords is at least $k+1$. But we're not given how many errors the code detects. So if the code detects only $1$ error than the minimal Hamming distance is $2$. But this is relevant for any code? How to do the given parameters come into play?
Best Answer
In this case, the blocks of the code are not of length $2n$ ($M$), they are of length $n$ ($M_1,M_2$), as the Hamming code is applied to each half separately. Thus the minimal Hamming distance between two words in the concatenated code is equal to that for just one of the halves, which is therefore $3$ since Hamming codes were used.