# Behaviour of an extended binary Hamming code when 3 errors have occurred

Take the [8,4] extended code for example. Note that extended binary Hamming codes are 3-error-detecting.

All single-bit errors are correctly decoded, while double-bit errors are detected but not correctable.

But how exactly does the code behave when the error is 3-bit? Does it always decode the received word to a wrong codeword (decoder error), or does it always detect but not correct such errors (i.e. treat 3-bit and 2-bit errors the same), or does it behave either way depending on the specific received word?

Thanks!

P.S. Here's the motivation for my above question: for a [7,4] (2-error-detecting) Hamming code, 2-bit errors will result in a decoder error, i.e. the received word will be decoded to a wrong codeword.

1. Words of the $[8,4,4]$ extended Hamming code. There are 16 of these, and they have even weight.
Added: As for decoding, suppose we use the parity check matrix $$H=\begin{bmatrix}1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0\\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0\end{bmatrix}$$ and regard the first seven bits as coming from the $[7,4,3]$ Hamming code with the eighth bit as parity check bit. Given received word $w,$ we compute $Hw,$ which is a four-bit word. If the received word is a codeword, $Hw$ will equal $0000.$ If the received word has odd weight, bit $1$ of $Hw$ will be set, and the decoder will assume one error. Under this assumption, if $Hw=1000$ then the error was in bit $8;$ otherwise, the last three bits of $Hw$ are the binary encoding of the position of the error. Finally, if the received word has even weight but is not a codeword, then $Hw$ will not equal $0000$ but its first bit will not be set. The decoder will then report a noncorrectable error.