Finding the overhead and distance of an unknown code based on message making algorithm

Question

for an information word M with m bits that is coded as following:

• M is coded into a word A using an unknown code that allows detection of not more than one error.
• the code word is the word obtained by a concatenation of A to itself. so A becomes AA and etc..

1)is it possible to know the overhead in the given code?

2)is it possible to know the distance of the code?

my attempt:

after thinking on it again and rereading the text, i began to realize(hopefully i am right now) that the question is about parity bit and not hamming. so if it can detect not more than one error, there's a parity bit, i think. based on that thought i tried to solve it:

1)I began to realize it's not about hamming code. if we can only detect not more than one error, then it's about the parity bit. so i think that if we add a bit to each byte of message, then the overhead will be 9 bytes total. so it can be calculated, i think.

2)since this sub problem is about the concatenation of A to itself, then like @YuvalFilmus said, the minimal distance should be twice now.

is it correct now? would really appreciate your comments and help.

thank you very much

Answer 1

The question is still pretty unclear, but I will provide a few hints and hope that they point you in the right direction.

The only code I'm aware of that can detect only one error and do nothing else is the repetition code \$R_2\$. This code takes a data bit \$b\$ and produces the codeword \$bb\$.

(Note that adding a parity bit can do more than detecting one error; it can detect any odd number of errors).

Now, by producing a code word that is a concatenation of \$A\$ with itself, all you're doing is using the same \$R_2\$ code again.

In summary, the mistery code is \$R_2\$, and the second code (which duplicates the word \$A\$) is just \$R_2\$ again.

The concatenation of two \$R_2\$ codes is equivalent to an \$R_4\$, where each bit is repeated four times; in other words, if the input is \$b\$, the output is \$bbbb\$. This code has distance \$4\$, can correct one error, and it can detect up to three errors.

Its code rate is \$1/4\$; it transmits four code bits per every data bit. The overhead then is \$3/4\$, or \$75\%\$.